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

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

", "advice": "

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

\n

The chain rule says that if $f(x)=g(h(x))$ then

\n

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

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One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

\n

Then 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}\\]

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

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This gives

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\\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}

\n

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}

\n

on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.

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

\n

$\\displaystyle \\frac{\\mathrm{d}f}{\\mathrm{d}x}=$ [[0]]

\n

Click on Show steps for more information. You will not lose any marks by doing so.

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The chain rule says that if $f(x)=g(h(x))$ then

\n

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

\n

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

\n

Then 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}\\]

\n

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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