// Numbas version: finer_feedback_settings {"name": "Chain rule - binomial", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Chain rule - binomial", "tags": ["Calculus", "calculus", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "Differentiation", "function of a function", "Steps", "steps"], "metadata": {"description": "
Differentiate $\\displaystyle (ax^m+b)^{n}$.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Differentiate the following function $f(x)$ using the chain rule.
", "advice": "$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$
\nThe chain rule says that if $f(x)=g(h(x))$ then
\n\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
\nOne way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
\nThen we use the chain rule in the form:
\n\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u}\\]
\nOnce you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
\nFor this example, we let $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.
\nThis gives
\n\\begin{align}
\\frac{\\mathrm{d}u}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m -1}} \\\\[1em]
\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u} &= \\simplify[std]{{n}u^{n-1}}
\\end{align}
Hence on substituting into the chain rule above we get:
\n\\begin{align}
\\frac{\\mathrm{d}f}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m-1} * ({n}*u^{n-1})} \\\\
&= \\simplify[std]{{m*a*n}x^{m-1}u^{n-1}} \\\\
&= \\simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}}
\\end{align}
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.
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\n$\\displaystyle \\frac{\\mathrm{d}f}{\\mathrm{d}x}=$ [[0]]
\nClick on Show steps for more information. You will not lose any marks by doing so.
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\n\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
\nOne way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
\nThen we use the chain rule in the form:
\n\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f}{\\mathrm{d}u}\\]
\nOnce you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.
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