// Numbas version: exam_results_page_options {"name": "Chain rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "name": "Chain rule", "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 $x$.

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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*}\\]

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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}/u)}\\\\\n \n &=& \\simplify[std]{{a*m}/({a}x+{b})}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x+{b}}$.

\n \n \n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "parts": [{"stepspenalty": 0.0, "prompt": "\n

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

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

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Click 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 \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}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

\n \n \n ", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "extensions": [], "statement": "

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

 

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1/08/2012:

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Added tags.

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Added description.

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Checked calculation. OK.

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Added information about Show steps. Altered to 0 marks lost rather than 1.

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Got rid of a redundant ruleset.

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Improved display in prompt.

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Checking range chosen so that the denominator of the result is never 0.

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Differentiate $\\displaystyle \\ln((ax+b)^{m})$

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