26 questions: Product Rule, Quotient Rule and Chain Rule. For those that want a thorough testing of their basic differentiation using the standard rules.
", "licence": "Creative Commons Attribution 4.0 International"}, "timing": {"timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "shufflequestions": false, "questions": [], "percentpass": 0.0, "allQuestions": true, "pickQuestions": 0, "type": "exam", "feedback": {"showtotalmark": true, "advicethreshold": 0.0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true, "enterreviewmodeimmediately": false, "showexpectedanswerswhen": "never", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "Chain rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["chain rule", "cosh", "derivatives of hyperbolic functions", "differential", "differentiate", "differentiating hyperbolic functions", "differentiation", "hyperbolic functions", "product rule", "sinh", "tanh"], "advice": "\n \n \nHere is a table of the derivatives of some of the hyperbolic functions:
\n \n \n \n| $f(x)$ | $\\displaystyle{\\frac{df}{dx}}$ |
|---|---|
| $\\sinh(bx)$ | $b\\cosh(bx)$ |
| $\\cosh(bx)$ | $b\\sinh(bx)$ |
| $\\tanh(bx)$ | $\\simplify{b*sech(bx)^2}$ |
a)
\n \n \n \n$f(x)=\\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}$
\n \n \n \nUse the product rule to obtain:
\\[\\frac{df}{dx} = \\simplify[std]{{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * Cosh({a1} * x + {b1})}\\]
b)
\n \n \n \n$f(x)=\\tanh(\\simplify[std]{{a}x+{b}})$
\n \n \n \nUsing the table above we get:
\\[\\frac{df}{dx} = \\simplify[std]{{a}*sech({a}x+{b})^2}\\]
c)
\n \n \n \n$f(x)=\\ln(\\cosh(\\simplify[std]{{a2}x+{b2}}))$
\n \n \n \nUsing the chain rule we find:
\n \n \n \n\\[\\frac{df}{dx} = \\simplify[std]{{a2} * tanh({a2} * x + {b2})}\\]
\n \n \n \n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n$f(x)=\\simplify[std]{ x ^ {n} * sinh({a1} * x + {b1})}$
\n$\\displaystyle{\\frac{df}{dx}=\\;\\;}$[[0]]
\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{n} * (x ^ {(n -1)}) * sinh({a1} * x + {b1}) + {a1} * (x ^ {n}) * Cosh({a1} * x + {b1})", "type": "jme"}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n$f(x)=\\tanh(\\simplify[std]{{a}x+{b}})$
\n$\\displaystyle{\\frac{df}{dx}=\\;\\;}$[[0]]
\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{a}*sech({a}x+{b})^2", "type": "jme"}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n$f(x)=\\ln(\\cosh(\\simplify[std]{{a2}x+{b2}}))$
\n$\\displaystyle{\\frac{df}{dx}=\\;\\;}$[[0]]
\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{a2} * tanh({a2} * x + {b2})", "type": "jme"}], "type": "gapfill", "marks": 0.0}], "statement": "\n \n \nWrite down the derivatives of the following functions $f(x)$ .
\n \n \n \nNote that in order to input the square of a function such as $\\sinh(x)$ you have to input it as $(\\sinh(x))^2$, similarly for the other hyperbolic functions.
\n \n \n \n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "b": {"definition": "random(-9..9)", "name": "b"}, "n": {"definition": "random(3..7)", "name": "n"}, "a1": {"definition": "random(-9..-1)", "name": "a1"}, "a2": {"definition": "random(2..9)", "name": "a2"}, "b1": {"definition": "random(1..9)", "name": "b1"}, "b2": {"definition": "random(-9..9)", "name": "b2"}}, "metadata": {"notes": "\n \t\t29/06/2012:
\n \t\tAdded and edited tags.
\n \t\t19/07/2012:
\n \t\tAdded description.
\n \t\tThere is also the problem of inputting functions of the form $xf(x)$ if $n=1$ or $2$ in the first question. So have reset $n$ to between $3$ and $7$. Otherwise would have to have an instruction here (perhaps depending on value of $n$).
\n \t\tChecked calculation.
\n \t\t23/07/2012:
\n \t\tAdded tags.
\n \t\t\n \t\t
Question appears to be working correctly.
\n \t\t1/08/2012:
\n \t\tThis is a copy of MAS114220122013CBA3_4 and is included in Diagnostic: Chain Rule Practice exam.
\n \t\t\n \t\t", "description": "
Differentiate the following functions: $\\displaystyle x ^ n \\sinh(ax + b),\\;\\tanh(cx+d),\\;\\ln(\\cosh(px+q))$
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Chain rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "chain rule", "derivative of a function of a function", "differentiation", "function of a function", "logarithm laws", "logarithms", "steps"], "advice": "\n \n \n$\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}$
First note that we can simplify this by using the rule that $\\simplify[std]{ln(a^r)=r*ln(a)}$.
Hence $\\simplify[std]{f(x) = ln(({a}x+{b})^{m})={m}ln({a}x+{b})}$
So we need to differentiate $\\simplify[std]{ln({a}x+{b})}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
For this example, we let $u=\\simplify[std]{{a}x +{b}}$ and we have $f(u)=\\simplify[std]{{m}*ln(u)}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a}}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{{m}/u} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a}) * ({m}/u)}\\\\\n \n &=& \\simplify[std]{{a*m}/({a}x+{b})}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x+{b}}$.
\\[\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [5.0, 6.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "({m*a})/({a}x+{b})", "type": "jme"}], "steps": [{"prompt": "\n \n \nThe chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
Differentiate the following function $f(x)$ using the chain rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "c": {"definition": "s2*random(1..9)", "name": "c"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "random(3..9)", "name": "m"}}, "metadata": {"notes": "\n \t\t\n \t\t
1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n \t\tGot rid of a redundant ruleset.
\n \t\tImproved display in prompt.
\n \t\tChecking range chosen so that the denominator of the result is never 0.
\n \t\t\n \t\t", "description": "
Differentiate $\\displaystyle \\ln((ax+b)^{m})$
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Chain rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "chain rule", "derivative of a function of a function", "differentiation", "function of a function", "steps"], "advice": "\n \n \n$\\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
For this example, we let $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{sqrt(u)=u^{1/2}}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{m*a}x ^ {m -1}}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{{1/2}*u^{-1/2}=1/(2*sqrt(u))} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{{m*a}x ^ {m-1} * (1/(2*sqrt(u)))}\\\\\n \n &=&\\simplify[std]{{m*a}x^{m-1}/(2*sqrt(u))}\\\\\n \n &=& \\simplify[std]{({a*m}x ^ {m-1})/(2*sqrt({a} * x^{m}+{b}))}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.
\\[\\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
\nInput all numbers as fractions or integers and not decimals.
\n\n ", "gaps": [{"notallowed": {"message": "
Input all numbers as fractions or integers and not decimals.
", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 1e-05, "vsetrange": [4.0, 5.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "({a*m}x ^ {m-1})/(2*sqrt({a} * x^{m}+{b}))", "type": "jme"}], "steps": [{"prompt": "\n \n \nThe chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
Differentiate the following function $f(x)$ using the chain rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "m": {"definition": "random(2..8)", "name": "m"}}, "metadata": {"notes": "\n \t\t1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK - but had to introduce more stringent accuracy constraints - see below.
\n \t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n \t\tGot rid of a redundant ruleset.
\n \t\tImproved display in prompt.
\n \t\tAdded decimal point to forbidden strings and included message not to input decimals.
\n \t\tIncreased accuracy threshold to abs diff of 0.00001 and tested the outcomes. OK.
\n \t\t", "description": "\n \t\tDifferentiate
\n \t\t\\[ \\sqrt{a x^m+b})\\]
\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Chain rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "chain rule", "derivative of a function of a function", "differentiation", "function of a function", "steps"], "advice": "\n \n \n$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
For this example, we let $u=\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{e^u}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a*m}x^{m-1} +{2*b}x}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{e^u} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x) * (e^u)}\\\\\n \n &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x)*e^({a}x^{m} +{b}x^2+{c})}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.
\\[\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "({m*a}x^{m-1}+{2*b}x)*e^({a}x^{m} +{b}x^2+{c})", "type": "jme"}], "steps": [{"prompt": "\n \n \nThe chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
Differentiate the following function $f(x)$ using the chain rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "c": {"definition": "s2*random(1..9)", "name": "c"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "random(3..4)", "name": "m"}}, "metadata": {"notes": "\n \t\t1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n \t\tGot rid of a redundant ruleset.
\n \t\tImproved display in prompt.
\n \t\t", "description": "Differentiate $\\displaystyle e^{ax^{m} +bx^2+c}$
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Chain rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "algebraic manipulation", "calculus", "chain rule", "derivative of the product of two functions", "differentiation", "product rule"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = ({a} * x^2+{b})^{n}} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {2*n*a}*x * ({a} * x^2+{b})^{n-1}}\\]
\n \n \n \nFor this last differentiation we used the chain rule.
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+x^{m} *{2*n*a}*x* ({a} * x^2+{b})^{n-1}}\\\\\n \n &=& \\simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+{2*n*a}*x^{m+1}* ({a} * x^2+{b})^{n-1}}\\\\\n \n &=& \\simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m}*({a}*x^2+{b})+{2*n*a}x^{2})} \\\\\n \n &=&\\simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m*a+2*a*n}*x^2+{m*b})}\n \n \\end{eqnarray*}\\]
\n \n \n \nHence $\\simplify[std]{g(x)={m*a+2*a*n}*x^2+{m*b}}$
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n$\\simplify[std]{f(x) = x ^ {m} * ({a} * x^2+{b})^{n}}$
The answer is in the form
\\[\\frac{df}{dx}=\\simplify[std]{x^{m-1}({a}x^2+{b})^{n-1}*g(x)}\\] for a polynomial $g(x)$.
You have to find $g(x)$.
\n$g(x)=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m*a+2*a*n}*x^2+{m*b}", "type": "jme"}], "steps": [{"prompt": "You should use the the product rule and the chain rule for this example.
", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "statement": "Differentiate the following function $f(x)$.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "m": {"definition": "random(3..9)", "name": "m"}, "n": {"definition": "random(3..9)", "name": "n"}}, "metadata": {"notes": "\n \t\t1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n \t\tGot rid of a redundant ruleset.
\n \t\t\n \t\t
\n \t\t", "description": "
The derivative of $\\displaystyle x ^ {m}(ax^2+b)^{n}$ is of the form $\\displaystyle x^{m-1}(ax^2+b)^{n-1}g(x)$. Find $g(x)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "derivative of a polynomial", "derivative of a product", "differentiation", "product rule", "steps"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = ({a} * x+{b})^{n}} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {n*a} * ({a} * x+{b})^{n-1}}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = {m}x ^ {m-1} * ({a} * x+{b})^{n}+{n*a}x^{m} * ({a} * x+{b})^{n-1}=x^{m-1}({a}x+{b})^{n-1}({m*a+n*a}x+{m*b})}\\]
So $\\simplify[std]{g(x)= {m*a+n*a}x+{m*b}}$
$\\simplify[std]{f(x) = x ^ {m} * ({a} * x+{b})^{n}}$
\n$g(x)=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m*a+n*a}x+{m*b}", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule. The answer will be of the form
\\[\\simplify[std]{x^{m-1}({a}x+{b})^{n-1}g(x)}\\] for a polynomial $g(x)$. You have to find $g(x)$
31/07/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tProblem with steps to be addressed. Now resolved.
\n \t\tClicking on Show steps does not lose any marks.
\n \t\t", "description": "\n \t\tDifferentiate $ x ^ m(ax+b)^n$ using the product rule. The answer will be of the form $x^{m-1}(ax+b)^{n-1}g(x)$ for a polynomial $g(x)$. Find $g(x)$.
\n \t\t\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "algebraic manipulation", "calculus", "derivatives", "differentiate a product", "differentiate polynomials", "differentiation", "elementary differentiation", "polynomials", "product rule", "steps"], "advice": "\n \n \n
The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = ({a} * x+{b})^{n}} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {n*a} * ({a} * x+{b})^{n-1}}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = {m}x ^ {m-1} * ({a} * x+{b})^{n}+{n*a}x^{m} * ({a} * x+{b})^{n-1}}\\]
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"pattern": "a/sqrt(b)", "result": "(sqrt(b)*a)/b"}]}, "parts": [{"stepspenalty": 0.0, "prompt": "$\\displaystyle \\simplify[std]{f(x) = x ^ {m} * ({a} * x+{b})^{n}}$
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClicking on Show steps gives you more information. You will not lose any marks by doing so.
", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m}x ^ {m-1} * ({a} * x+{b})^{n}+{n*a}x^{m} * ({a} * x+{b})^{n-1}", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "m": {"definition": "random(3..9)", "name": "m"}, "n": {"definition": "random(3..9)", "name": "n"}}, "metadata": {"notes": "\n \t\t31/07/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tSteps problem to be addressed via an issue. Now resolved.
\n \t\tChecked calculation.OK.
\n \t\tImproved prompt display.
\n \t\tClicking on Show steps does not lose any marks.
\n \t\t", "description": "Differentiate $f(x) = x^m(a x+b)^n$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "algebraic manipulation", "calculus", "derivatives", "differentiating a product", "differentiating square roots", "differentiation", "elementary differentiation", "product rule", "steps"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = sqrt({a} * x+{b})} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {a}/2* ({a} * x+{b})^{-1/2}}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\begin{eqnarray*} \\frac{df}{dx}&=& \\simplify[std]{{m}x ^ {m-1} * sqrt({a} * x+{b})+{a/2}x^{m} * ({a} * x+{b})^{-1/2}}\\\\\n \n &=& \\simplify[std]{{m}x ^ {m-1} * sqrt({a} * x+{b})+{a}x^{m}/(2*sqrt({a} * x+{b}))}\\\\\n \n &=& \\simplify[std]{(2*{m}x^{m-1}({a}x+{b})+ {a}x^{m})/(2*sqrt({a} * x+{b}))}\\\\\n \n &=&\\simplify[std]{(x^{m-1}({2*m}({a}x+{b})+{a}x))/(2*sqrt({a} * x+{b}))}\\\\\n \n &=&\\simplify[std]{x^{m-1}/(2*sqrt({a} * x+{b}))({2*m*a+a}x+{2*m*b})}\n \n \\end{eqnarray*}\\]
Hence $g(x)=\\simplify[std]{{2*m*a+a}x+{2*m*b}}$
$\\simplify[std]{f(x) = x ^ {m} * sqrt({a} * x+{b})}$
\nThe answer is in the form \\[\\frac{df}{dx}=\\simplify[std]{ x^{m-1}/(2*sqrt({a}x+{b}))g(x)}\\]
for a polynomial $g(x)$. You have to find $g(x)$.
$g(x)=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{2*m*a+a}x+{2*m*b}", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": 1.0, "name": "a"}, "b": {"definition": "random(1..9)", "name": "b"}, "m": {"definition": "random(2..9)", "name": "m"}}, "metadata": {"notes": "\n \t\t31/07/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tSteps problem to be addressed. Now resolved.
\n \t\tChecked calculation.OK.
\n \t\tImproved prompt display.
\n \t\tClicking on Show steps does not lose any marks.
\n \t\t", "description": "Differentiate $ x ^m \\sqrt{a x+b}$.
The answer is in the form $\\displaystyle \\frac{x^{m-1}g(x)}{2\\sqrt{ax+b}}$
for a polynomial $g(x)$. Find $g(x)$.
The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = ({a} + {b} * x) ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = e ^ ({n} * x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {n} * e ^ ({n} * x)}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) }\\]
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"pattern": "a/sqrt(b)", "result": "(sqrt(b)*a)/b"}]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n$\\simplify[std]{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x)", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(1..9)", "name": "a"}, "b": {"definition": "s1*random(2..9)", "name": "b"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "random(2..8)", "name": "m"}, "n": {"definition": "s2*random(2..5)", "name": "n"}}, "metadata": {"notes": "\n \t\t31/07/2012:
\n \t\tAdded tags.
\n \t\tImproved display of prompt.
\n \t\tChecked calculation.
\n \t\tAllowed no penalty on looking at Steps.
\n \t\tIssue with Show steps to be resolved. Has been resolved.
\n \t\t", "description": "Differentiate the function $(a + b x)^m e ^ {n x}$ using the product rule.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "differentiating a product", "differentiating trigonometric functions", "differentiation", "product rule", "steps", "trigonometric functions"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = ({a} + {b} * x) ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = sin({n} * x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {n} * cos({n} * x)}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = {m*b}({a} + {b} * x) ^ {m-1} * sin({n} * x)+{n}*({a} + {b} * x) ^ {m} * cos({n} * x)}\\]
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"pattern": "a/sqrt(b)", "result": "(sqrt(b)*a)/b"}]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n$\\simplify[std]{f(x) = ({a} + {b} * x) ^ {m} * sin({n} * x)}$
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m*b}({a} + {b} * x) ^ {m-1} * sin({n} * x)+{n}*({a} + {b} * x) ^ {m} * cos({n} * x)", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(1..4)", "name": "a"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "b": {"definition": "s1*random(1..5)", "name": "b"}, "m": {"definition": "random(2..8)", "name": "m"}, "n": {"definition": "random(2..6)", "name": "n"}}, "metadata": {"notes": "\n \t\t31/07/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tSteps problem to be addressed. Now resolved.
\n \t\tChecked calculation.OK.
\n \t\tImproved prompt display.
\n \t\tClicking on Show steps not lose any marks.
\n \t\t", "description": "Differentiate $ (a+bx) ^ {m} \\sin(nx)$
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "derivative of a product", "differentiating a product", "differentiating trigonometric functions", "differentiation", "product rule", "steps", "trigonometric functions"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = cos({a} * x+{b})} \\Rightarrow \\simplify[std]{Diff(v,x,1) = -{a} * sin({a} * x+{b})}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = {m}x ^ {m-1} * cos({a} * x+{b})-{a}x^{m} * sin({a} * x+{b})}\\]
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"pattern": "a/sqrt(b)", "result": "(sqrt(b)*a)/b"}]}, "parts": [{"stepspenalty": 0.0, "prompt": "$\\simplify[std]{f(x) = x ^ {m} * cos({a} * x+{b})}$
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClicking on Show steps gives you more information. You will not lose any marks by doing so.
", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m}x ^ {m-1} * cos({a} * x+{b})-{a}x^{m} * sin({a} * x+{b})", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "m": {"definition": "random(2..9)", "name": "m"}}, "metadata": {"notes": "\n \t\t31/07/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tSteps problem to be addressed. Now resolved.
\n \t\tChecked calculation.OK.
\n \t\tImproved prompt display.
\n \t\tClicking on Show steps does not lose any marks.
\n \t\t", "description": "Differentiate $x^m\\cos(ax+b)$
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "calculus", "derivative of a product", "differentiating a product", "differentiating exponential functions", "differentiating trigonometric functions", "differentiation", "product rule", "steps"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = sin({a} + {b} * x)}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {b} * cos({a} + {b} * x)}\\]
\n \n \n \n\\[\\simplify[std]{v = e ^ ({n} * x)} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {n} * e ^ ({n} * x)}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{{b} * cos({a} + {b} * x) * e ^ ({n} * x) + {n} * sin({a} + {b} * x) * e ^ ({n} * x)}\\\\\n \n &=&\\simplify[std]{({b}cos({a}+{b}x)+{n}sin({a}+{b}x))e^({n}x)}\n \n \\end{eqnarray*}\\]
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"pattern": "a/sqrt(b)", "result": "(sqrt(b)*a)/b"}]}, "parts": [{"stepspenalty": 0.0, "prompt": "$\\simplify[std]{f(x) = sin({b}x + {a}) * e ^ ({n} * x)}$
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{b} * cos({a} + {b} * x) * e ^ ({n} * x) + {n} * sin({a} + {b} * x) * e ^ ({n} * x)", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(1..9)", "name": "a"}, "b": {"definition": "s1*random(2..9)", "name": "b"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "random(2..8)", "name": "m"}, "n": {"definition": "s2*random(2..5)", "name": "n"}}, "metadata": {"notes": "\n \t\t31/07/2012:
\n \t\tChecked calculation.
\n \t\tAdded tags.
\n \t\tAllowed no penalty on looking at Show steps.
\n \t\t", "description": "Differentiate $ \\sin(ax+b) e ^ {nx}$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "algebraic manipulation", "calculus", "derivative of a product", "differentiation", "polynomials", "product rule", "steps"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = ({a}x+{b}) ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m*a}({a}x+{b}) ^ {m -1}}\\]
\n \n \n \n\\[\\simplify[std]{v = ({c} * x+{d})^{n}} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {n*c} * ({c} * x+{d})^{n-1}}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = {m*a}({a}x+{b}) ^ {m-1} * ({c} * x+{d})^{n}+{n*c}({a}x+{b})^{m} * ({c} * x+{d})^{n-1}=({a}x+{b})^{m-1}({c}x+{d})^{n-1}({m*a*c+n*c*a}x+{m*a*d+n*c*b})}\\]
on taking out the common terms $\\simplify[std]{({a}x+{b})^{m-1}({c}x+{d})^{n-1}}$.
So $\\simplify[std]{g(x)= {m*a*c+n*c*a}x+{m*a*d+n*c*b}}$
$\\simplify[std]{f(x) = ({a} * x+{b})^{m}*({c}x+{d})^{n}}$
\n$g(x)=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{m*a*c+n*c*a}x+{m*a*d+n*c*b}", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule. The answer will be of the form
\\[\\simplify[std]{({a}x+{b})^{m-1}({c}x+{d})^{n-1}g(x)}\\] for a polynomial $g(x)$. You have to find $g(x)$
31/07/2012:
\n \t\tAdded tags.
\n \t\tCorrected mistake in statement.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tProblem with steps to be addressed. Now resolved.
\n \t\tClicking on Show steps does not lose any marks.
\n \t\t", "description": "Differentiate $ (ax+b)^m(cx+d)^n$ using the product rule. The answer will be of the form $(ax+b)^{m-1}(cx+d)^{n-1}g(x)$ for a polynomial $g(x)$. Find $g(x)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Product Rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Steps", "algebraic manipulation", "derivative", "deriving a function", "differentiate", "differentiating a function", "differentiating a product of functions", "differentiation", "exponential function", "functions", "product rule", "steps"], "advice": "\n \n \nThe product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
For this example:
\n \n \n \n\\[\\simplify[dPoly]{u = ({a} + {b} * x) ^ {m}}\\Rightarrow \\simplify[dPoly]{Diff(u,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1}}\\]
\n \n \n \n\\[\\simplify{v = e ^ ({n} * x)} \\Rightarrow \\simplify{Diff(v,x,1) = {n} * e ^ ({n} * x)}\\]
\n \n \n \nHence on substituting into the product rule above we get:
\n \n \n \n\\[\\simplify[dPoly]{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x) = ({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}\\]
\n \n \n \nThe last step was to take out the common term $\\simplify[dPoly]{({a} + {b} * x) ^ {m -1} * e ^ ({n} * x)}$.
\n \n \n \nHence \\[\\simplify[dPoly]{g(x) = {m * b + n * a} + {n * b} * x}\\].
\n \n \n \n ", "rulesets": {"std": ["all", "!collectNumbers"], "dpoly": ["std", "fractionNumbers"]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n$\\simplify[dPoly]{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$
\nYou are given that \\[\\simplify[dPoly]{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}\\]
\nfor a polynomial $g(x)$. You have to find $g(x)$.
\n$g(x)=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
\n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "dPoly", "marks": 3.0, "answer": "({((m * b) + (n * a))} + ({(n * b)} * x))", "type": "jme"}], "steps": [{"prompt": "The product rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]
Differentiate the following function $f(x)$ using the product rule.
", "variable_groups": [], "progress": "testing", "type": "question", "variables": {"a": {"definition": "random(1..4)", "name": "a"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "b": {"definition": "s1*random(1..5)", "name": "b"}, "m": {"definition": "random(2..8)", "name": "m"}, "n": {"definition": "random(2..6)", "name": "n"}}, "metadata": {"notes": "\n \t\t20/06/2012:
\n \t\tAdded tags.
\n \t\t4/7/2012:
Added tags.
\n \t\t31/07/2012:
\n \t\tChecked calculation.
\n \t\tAllowed no penalty on looking at Steps.
\n \t\t", "description": "Differentiate the function $f(x)=(a + b x)^m e ^ {n x}$ using the product rule. Find $g(x)$ such that $f\\;'(x)= (a + b x)^{m-1} e ^ {n x}g(x)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Quotient rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "algebraic manipulation", "calculus", "derivative of a quotient", "differentiation", "quotient rule", "steps"], "advice": "\n \n \nThe quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
For this example:
\n \n \n \n\\[\\simplify[std]{u = ({a}x+{b})}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {a}}\\]
\n \n \n \n\\[\\simplify[std]{v = ({c} * x^2+{d}x+{f})} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {2*c}x+{d}}\\]
\n \n \n \nHence on substituting into the quotient rule above we get:
\n \n \n \n\\[\\begin{eqnarray*} \\frac{df}{dx}&=&\\simplify[std]{({a}({c}x^2+{d}x+{f})-({2*c}x+{d})({a}x+{b}))/({c}x^2+{d}x+{f})^2}\\\\\n \n &=&\\simplify[std]{({a*c}x^2+{a*d}x+{a*f}-{2*c*a}x^2-{a*d+2*c*b}x+{d*b})/({c}x^2+{d}x+{f})^2}\\\\\n \n &=&\\simplify[std]{({-c*a}x^2+{-2*b*c}x+{a*f-d*b})/({c}x^2+{d}x+{f})^2}\n \n \\end{eqnarray*}\\]
Hence $g(x)=\\simplify[std]{{-c*a}x^2+{-2*b*c}x+{a*f-d*b}}$
\\[\\simplify[std]{f(x) = ({a} * x+{b})/({c}x^2+{d}x+{f})}\\]
You are given that \\[\\frac{df}{dx}=\\simplify[std]{g(x)/({c}x^2+{d}x+{f})^2}\\]
for a polynomial $g(x)$. You are asked to find $g(x)$
$g(x)=\\;$[[0]]
\nInput numbers as fractions or integers and not as decimals.
\nClick on Show steps for more information. You will not lose any marks by doing so.
\n ", "gaps": [{"notallowed": {"message": "Input numbers as fractions or integers and not as decimals.
", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "{-c*a}x^2+{-2*b*c}x+{a*f-b*d}", "type": "jme"}], "steps": [{"prompt": "The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1) = (v * Diff(u,x,1) - u * Diff(v,x,1))/v^2}\\]
Differentiate the following function $f(x)$ using the quotient rule.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "c": {"definition": "if(a*d=b*c1,c1+1,c1)", "name": "c"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "d": {"definition": "s2*random(1..9)", "name": "d"}, "f": {"definition": "random(-9..9)", "name": "f"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "det": {"definition": "a*f-b*d", "name": "det"}, "c1": {"definition": "random(1..8)", "name": "c1"}}, "metadata": {"notes": "\n \t\t1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n \t\tChanged std rule set to include !noLeadingMinus, so polynomials don't change order. Got rid of a redundant ruleset.
\n \t\tImproved display in various places.
\n \t\tAdded condition that numbers have to be inout as fractions or integers - added decimal point to forbidden strings.
\n \t\t\n \t\t", "description": "
The derivative of $\\displaystyle \\frac{ax+b}{cx^2+dx+f}$ is $\\displaystyle \\frac{g(x)}{(cx^2+dx+f)^2}$. Find $g(x)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Quotient rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "Steps", "algebraic manipulation", "calculus", "derivative of a quotient", "differentiation", "quotient rule", "steps"], "advice": "\n \n \nThe quotient rule says that if $u$ and $v$ are functions of $x$ then
\n \n \n \n\\[\\simplify[std]{Diff(u/v,x,1)=(v * Diff(u,x,1) -(u * Diff(v,x,1))) / v ^ 2}\\]
\n \n \n \nFor this example:
\n \n \n \n\\[\\simplify[std]{u = {a} * x + {b}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {a}}\\]
\n \n \n \n\\[\\simplify[std]{v = Sqrt({c} * x + {d})} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {c} / (2 * Sqrt({c} * x + {d}))}\\]
\n \n \n \nHence on substituting into the quotient rule above we get:
\n \n \n \n\\[\\simplify[std]{Diff(f,x,1) = ({a} * Sqrt({c} * x + {d}) -(({a} * x + {b}) * Diff(v,x,1))) / ({c} * x + {d}) = ({a} * Sqrt({c} * x + {d}) -(({c} * ({a} * x + {b})) / (2 * Sqrt({c} * x + {d})))) / ({c} * x + {d}) = ({2 * a} * ({c} * x + {d}) -({c} * ({a} * x + {b}))) / (2 * ({c} * x + {d}) ^ (3 / 2)) = ({a * c} * x + {2 * a * d -(c * b)}) / (2 * ({c} * x + {d}) ^ (3 / 2))}\\]
\n \n \n \nHence \\[\\simplify[std]{g(x) = {a * c} * x + {2 * a * d -(c * b)}}\\].
\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n\\[\\simplify[std]{f(x) = ({a} * x + {b}) / Sqrt({c} * x + {d})}\\]
\nYou are given that \\[\\simplify[std]{Diff(f,x,1) = g(x) / (2 * ({c} * x + {d}) ^ (3 / 2))}\\]
\nfor a polynomial $g(x)$. You have to find $g(x)$.
\nInput all numbers as fractions or integers.
\nYou can click on Show steps to get help. You will not lose any marks if you do so.
\n$g(x)=\\;$[[0]]
\n ", "gaps": [{"notallowed": {"message": "Input all numbers as fractions or integers.
", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "all", "marks": 3.0, "answer": "(({(a * c)} * x) + {((2 * a * d) + ( - (c * b)))})", "type": "jme"}], "steps": [{"prompt": "The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u/v,x,1)=(v * Diff(u,x,1) -(u * Diff(v,x,1))) / v ^ 2}\\]
Differentiate the following function $f(x)$ using the quotient rule or otherwise.
", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(1..8)", "name": "a"}, "c": {"definition": "random(1,3,5,7)", "name": "c"}, "b": {"definition": "if(2|a,random(-7..7#2),random(-8..8#2))", "name": "b"}, "d": {"definition": "if(a*d1=b*c,abs(d1)+1,d1)", "name": "d"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "d1": {"definition": "s1*random(1..8)", "name": "d1"}}, "metadata": {"notes": "\n \t\t1/08/2012:
\n \t\tAdded tags.
\n \t\tAdded description.
\n \t\tChecked calculation. OK.
\n \t\tAdded information about Show steps. Altered to 0 marks lost rather than 1.
\n \t\tChanged std rule set to include !noLeadingMinus, so polynomials don't change order. Got rid of a redundant ruleset.
\n \t\tImproved display in various places.
\n \t\tAdded condition that numbers have to be input as fractions or integers - added decimal point to forbidden strings.
\n \t\t", "description": "The derivative of $\\displaystyle \\frac{ax+b}{\\sqrt{cx+d}}$ is $\\displaystyle \\frac{g(x)}{2(cx+d)^{3/2}}$. Find $g(x)$.
", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "extensions": [], "custom_part_types": [], "resources": []}