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If we have a function of the form $y=f(g(x))$, sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:

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\\[ \\frac{dy}{dx} = \\frac{du}{dx} \\times \\frac{dy}{du}.\\]

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This can be split up into stages:

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Calculate the derivative of $y=\\simplify[all,fractionNumbers]{{k}({a}*x^{m}+{b})^{n}}$.

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Following the process of calculating the derivative of a function using the chain rule, we must first identify $g(x)$. Since the function is of the form $y=f(g(x))$, we are looking for the 'inner' function:

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So, for $y=\\simplify[all,fractionNumbers]{{k}({a}*x^{m}+{b})^{n}}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}x^{m}+{b}}.\\]

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If we now set $u=g(x)$, we can rewrite $y$ in terms of $u$ such that $y=f(u)$:

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\\[y=\\simplify[all, fractionNumbers,unitFactor]{{k}u^{n}}.\\]

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Next, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:

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\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{{a*m}x^{m-1}}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{{k*n}*u^{n-1}}.\\]

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Plugging these into the chain rule:

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\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=\\simplify[all,fractionNumbers]{{a*m}x^{m-1}} \\times\\simplify[all, fractionNumbers, unitFactor]{{k*n}u^{n-1}}, \\\\ &=\\simplify[all, fractionNumbers, unitFactor]{{k*n*a*m}*x^{m-1}u^{n-1}}. \\end{split} \\]

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Finally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ terms using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{{a}x^{m}+{b}}$:

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\\[ \\frac{dy}{dx} = \\simplify[all, fractionNumbers, unitFactor]{{k*n*a*m}*x^{m-1}({a}x^{m}+{b})^{n-1}}.\\]

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$\\frac{dy}{dx}=$[[0]]

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