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See below for a worked solution of this particular problem.

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You might find this booklet from Mathcentre useful for revising the chain rule for differentiation.

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

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Formally the chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
or
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]

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(Note that $h(x)$ is equivalent to $u$.)

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For this example, let $u=\\simplify[std]{{a} * x^{m}+{b}}$. We then have $f(u)=\\simplify[std]{u^{n}}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{m*a}x ^ {m -1}}\\\\ \\frac{df(u)}{du} &=& \\simplify[std]{{n}u^{n-1}} \\end{eqnarray*}\\]

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Hence:

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\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\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{eqnarray*}\\]

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(Remember that your final answer should be in terms of $x$ not $u$.)

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Informally you can think:

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$f(x) = (something)^{\\simplify[std]{{n}}}$.

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So $\\frac{df}{dx} = \\simplify[std]{{n}}(something)^{\\simplify{{n}-1}} \\times \\frac{d}{dx} (something)$

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

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

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Differentiate the following function using the chain rule.

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

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