11 Questions on the chain rule.
", "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}, "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", "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", "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", "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": "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. (Video).", "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", "video"], "advice": "\n \n \n$\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}$
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 $u=\\simplify[std]{e^({a}x) +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{cos(u)}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a}e^({a}x) +{2*b}x}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{-sin(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}e^({a}x) +{2*b}x) * (-sin(u))}\\\\\n \n &=& \\simplify[std]{-({a}e^({a}x) +{2*b}x)*sin(e^({a}x) +{b}x^2+{c})}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{e^({a}x) +{b}x^2+{c}}$.
\\[\\simplify[std]{f(x) = cos(e^({a}x) +{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.
\nYou will also find a video solving a similar example in Show steps.
", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 3.0, "answer": "-({a}e^({a}x)+{2*b}x)*sin(e^({a}x) +{b}x^2+{c})", "type": "jme"}], "steps": [{"prompt": "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}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
The following video sets out the solution for a similar example.
\n \n", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "statement": "
Differentiate the following function $f(x)$ using the chain rule.
\n", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..9)", "name": "a"}, "s2": {"definition": "random(1,-1)", "name": "s2"}, "c": {"definition": "s2*random(1..9)", "name": "c"}, "b": {"definition": "s1*random(1..9)", "name": "b"}, "s1": {"definition": "random(1,-1)", "name": "s1"}}, "metadata": {"notes": "\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\t", "description": "Differentiate $\\displaystyle \\cos(e^{ax}+bx^2+c)$.
\nContains a video solving a similar example.
", "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": []}