// Numbas version: exam_results_page_options {"name": "Harry's copy of cormac's copy of Chain rule - exponential of polynomial, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Harry's copy of cormac's copy of Chain rule - exponential of polynomial, ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "variables": {"s2": {"name": "s2", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(1,-1)"}, "s1": {"name": "s1", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(1,-1)"}, "c": {"name": "c", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "s2*random(1..9)"}, "a": {"name": "a", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(2..9)"}, "m": {"name": "m", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "random(3..4)"}, "b": {"name": "b", "group": "Ungrouped variables", "templateType": "anything", "description": "", "definition": "s1*random(1..9)"}}, "preamble": {"js": "", "css": ""}, "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m"], "statement": "

Differentiate the following function $f(x)$ using the chain rule.

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\\[\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}\\]

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$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

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$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
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 $x$.

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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*}\\]

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Hence on substituting into the chain rule above we get:

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\\[\\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 $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.

\n\t \n\t \n\t", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Differentiate $\\displaystyle e^{ax^{m} +bx^2+c}$

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