// Numbas version: finer_feedback_settings {"name": "Philip'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": [{"metadata": {"description": "\n\t\t
Differentiate
\n\t\t\\[ \\sqrt{a x^m+b})\\]
\n\t\t", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "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 $x$.
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*}\\]
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]{{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 $\\simplify[std]{{a}x^{m}+{b}}$.
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 $x$.
\\[\\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}\\]
\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\tInput all numbers as fractions or integers and not decimals.
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Input all numbers as fractions or integers and not decimals.
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", "variable_groups": [], "type": "question", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Johnny Yi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2810/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}, {"name": "Philip Charlton", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3474/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Johnny Yi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2810/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}, {"name": "Philip Charlton", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3474/"}]}