Differentiate $\\displaystyle (ax^m+b)^{n}$.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "\n \n \n$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$
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 $f(u)=\\simplify[std]{u^{n}}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{m*a}x ^ {m -1}}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{{n}u^{n-1}} \\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} * ({n}*u^{n-1})}\\\\\n \n &=&\\simplify[std]{{m*a*n}x^{m-1}u^{n-1}}\\\\\n \n &=& \\simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.
\\[\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}\\]
\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "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 $\\displaystyle e^{ax^{m} +bx^2+c}$
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "\n\t \n\t \n\t$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
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\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{e^u} \\end{eqnarray*}\\]
Hence on substituting into the chain rule above we get:
\n\t \n\t \n\t \n\t\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x) * (e^u)}\\\\\n\t \n\t &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x)*e^({a}x^{m} +{b}x^2+{c})}\n\t \n\t \\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\t\t\t$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\n\t\t\tClick on Show steps for more information. You will not lose any marks by doing so.
\n\t\t\t", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\tThe 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 $\\displaystyle \\ln((ax+b)^{m})$
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "$\\simplify[std]{f(x) = ln({a}x+{b}x^{m})}$
\nThe chain rule applies to $f(x)=g(h(x))$ where
\n\\[ g(h) = \\ln(h) {\\rm~and~} \\simplify[std]{h(x) = {a}x+{b}x^{m}}.\\]
\nThen we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{dh}{dx} \\cdot \\frac{dg}{dh}\\]
Calculate the derivative of $h(x)$ and $g(h)$: \\[\\frac{dh}{dx} = \\simplify[std]{{a}+{b*m}x^{m-1}}\\]
\n\\[\\frac{dg}{dh} = \\frac{1}{h}\\]
\nHence on substituting $h = h(x) = \\simplify[std]{{a}x+{b}x^{m}}$ we finally have
\n\\[\\frac{df}{dx} = \\simplify[std]{({a}x+{b*m}x^{m-1})/({a}x+{b}x^{m})} \\]
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\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 3, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "({a}+{b*m}x^{m-1})/({a}x+{b}x^{m})", "answerSimplification": "std", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [5, 6], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "7.4 Differentiate product of sin/cos/ln/e function and quadratic", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Musa Mammadov", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4417/"}, {"name": "Simon James", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18202/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the product rule. Be sure to use brackets where required, e.g. \"e^(2x)\", \"sin(2x)\", \"ln(2x)\".
", "advice": "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 we can set
\n\\[ u = \\var{quadratic}, \\quad v = \\var{other} \\]
\nDifferentiating these gives
\n\\[ \\simplify[std]{Diff(u,x,1)} = \\var{quad_deriv}, \\quad \\simplify[std]{Diff(v,x,1)} = \\var{other_deriv} \\]
\nAnd so combining into our rule then leads to
\n\\[ \\simplify[std]{ u * Diff(v,x,1) + v * Diff(u,x,1)} = \\var{prod_deriv} \\]
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\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "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 quotient rule. Be sure to use brackets where required, e.g. \"e^(2x)\", \"sin(2x)\", \"ln(2x)\".
", "advice": "The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u / v,x,1)} = \\frac{\\simplify[std]{v * Diff(u,x,1)} -\\simplify[std]{ u * Diff(v,x,1) } }{v^2}\\]
For this example we can set
\n\\[ u = \\var{quadratic}, \\quad v = \\var{other} \\]
\n\\[ u = \\var{other}, \\quad v = \\var{quadratic} \\]
\nDifferentiating these gives
\n\\[ \\simplify[std]{Diff(u,x,1)} = \\var{quad_deriv}, \\quad \\simplify[std]{Diff(v,x,1)} = \\var{other_deriv} \\]
\n\\[ \\simplify[std]{Diff(u,x,1)} = \\var{other_deriv}, \\quad \\simplify[std]{Diff(v,x,1)} = \\var{quad_deriv} \\]
\nAnd so combining into our rule then leads to
\n\\[ \\frac{\\simplify[std]{ v * Diff(u,x,1) - u * Diff(v,x,1)}}{v^2} = \\var{quotient_deriv} \\]
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\n$\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\nClicking on Show steps gives you more information, you will not lose any marks by doing so.
", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "The quotient rule says that if $u$ and $v$ are functions of $x$ then
\\[\\simplify[std]{Diff(u / v,x,1)} = \\frac{\\simplify[std]{v * Diff(u,x,1)} -\\simplify[std]{ u * Diff(v,x,1) } }{v^2}\\]
Finding the coordinates and determining the nature of the stationary points on a polynomial function
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"y12": {"name": "y12", "group": "Ungrouped variables", "definition": "2(x12^3)-3(x12+x22)*x12^2+6*x12*x22*x12+c02", "description": "", "templateType": "anything", "can_override": false}, "y03": {"name": "y03", "group": "Ungrouped variables", "definition": "random(-10..10)", "description": "", "templateType": "anything", "can_override": false}, "y32": {"name": "y32", "group": "Ungrouped variables", "definition": "if(y12\\[ \\simplify[all,!noLeadingMinus,canonicalOrder]{y = {pm}*(2x^3-3{(x12+x22)}x^2+6{x12*x22}x+{c02})} \\]
\n\n
Determine the coordinates and the nature of the stationary points using the following steps.
\n\n
First Derivative
\n$y^{\\prime}(x) =$ [[0]]
\n\n
Give the values of x where stationary points occur:
\nsmallest-$x_1$ =[[1]]
\nlargest-$x_2$ = [[2]]
\n\n
Use the sign test to classify these as a local maximum, local minimum, or point of inflection.
\nBelow $x_1$: [[3]]
\nBetween $x_1$ and $x_2$: [[4]]
\nAbove $x_2$: [[5]]
\n\n
Therefore, $x_1$ is a [[6]] and $x_2$ is a [[7]]
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false, "startpassword": "", "autoSubmit": true, "allowAttemptDownload": false, "downloadEncryptionKey": "", "showresultspage": "oncompletion"}, "timing": {"allowPause": false, "timeout": {"action": "warn", "message": "Time has run out.
"}, "timedwarning": {"action": "warn", "message": "You have 5 minutes before time runs out.
"}}, "feedback": {"enterreviewmodeimmediately": true, "showactualmarkwhen": "inreview", "showtotalmarkwhen": "inreview", "showanswerstatewhen": "inreview", "showpartfeedbackmessageswhen": "inreview", "showexpectedanswerswhen": "inreview", "showadvicewhen": "inreview", "allowrevealanswer": false, "intro": "Instructions:
\n| 1. | \nComplete the questions within 90 minutes and achieve 80% or higher. | \n
| 2. | \nYou can take this self-assessment as many times as you need, until you receive a satisfactory grade (you don't need to achieve 80% when you 'give it a go' the first time) | \n
| 3. | \nUse the \"Print this results summary\" and save as a pdf after you complete your attempt. You will need the printout showing all questions for your module submission. | \n
Congratulations! You have achieved the minimum threshold for this module's self-assessment.
\nUse the \"Print this results summary\" and save your attempt as a pdf. You will need the printout showing all questions for your module submission.
", "threshold": "80"}, {"message": "Unfortunately you have not achieved the minimum score.
\nIf this is your first 'Give it a go' attempt - don't despair! This is exactly why we take our first attempt - to see how we're going and whether we need more practice in order to complete the quest.
\nYou should still use the \"Print this results summary\" option to save a copy of your results as a pdf, which will help with your learning and can also be shared with your tutors so they can help with certain questions.
\n