// Numbas version: exam_results_page_options {"name": "Harry's copy of Chain rule - square root of polynomial, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "\n\t \n\t \n\t

$\\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}$
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 $x$.

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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\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{{1/2}*u^{-1/2}=1/(2*sqrt(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]{{m*a}x ^ {m-1} * (1/(2*sqrt(u)))}\\\\\n\t \n\t &=&\\simplify[std]{{m*a}x^{m-1}/(2*sqrt(u))}\\\\\n\t \n\t &=& \\simplify[std]{({a*m}x ^ {m-1})/(2*sqrt({a} * x^{m}+{b}))}\n\t \n\t \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.

\n\t \n\t \n\t", "type": "question", "parts": [{"type": "gapfill", "marks": 0, "showCorrectAnswer": true, "scripts": {}, "stepsPenalty": 0, "steps": [{"marks": 0, "scripts": {}, "type": "information", "showCorrectAnswer": true, "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t

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\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t"}], "prompt": "\n\t\t\t

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

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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.

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Input all numbers as fractions or integers and not decimals.

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\n\t\t\t", "gaps": [{"type": "jme", "marks": 3, "checkingtype": "absdiff", "vsetrangepoints": 5, "notallowed": {"strings": ["."], "showStrings": false, "message": "

Input all numbers as fractions or integers and not decimals.

", "partialCredit": 0}, "vsetrange": [4, 5], "answersimplification": "std", "expectedvariablenames": [], "answer": "({a*m}x ^ {m-1})/(2*sqrt({a} * x^{m}+{b}))", "showpreview": true, "checkingaccuracy": 1e-05, "scripts": {}, "showCorrectAnswer": true, "checkvariablenames": false}]}], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "tags": ["Calculus", "MAS1601", "SFY0004", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function"], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "\n\t\t

Differentiate

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\\[ \\sqrt{a x^m+b})\\]

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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 - but had to introduce more stringent accuracy constraints - see below.

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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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Added decimal point to forbidden strings and included message not to input decimals.

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Increased accuracy threshold to abs diff of 0.00001 and tested the outcomes. OK.

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

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