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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 $x$.
\n\t \n\t \n\t \n\t 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*}\\]
\n\t \n\t \n\t \n\t 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 $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.
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\n\t\t\t $\\displaystyle \\frac{df}{dx}=\\;$[[0]]
\n\t\t\t Click on Show steps for more information. You will not lose any marks by doing so.
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\\[\\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 $x$.
\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t"}], "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": "({m*a}x^{m-1}+{2*b}x)*e^({a}x^{m} +{b}x^2+{c})", "answerSimplification": "std", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "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", "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/"}], "resources": []}]}], "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/"}]}