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\$$f(x)=({\\var{a2}x^{\\var{a3}}+\\var{a4}})^\\var{a1}\$$

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Recall the chain rule:   \$$\\frac{df}{dx}=\\frac{df}{du}.\\frac{du}{dx}\$$

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let \$$u=\\var{a2}x^{\\var{a3}}+\\var{a4}\$$    then   \$$f(x)=u^\\var{a1}\$$

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\$$\\frac{df}{du}=\\var{a1}u^{\\var{a1}-1}\$$  and  \$$\\frac{du}{dx}=\\var{a3}*\\var{a2}x^{\\var{a3}-1}\$$

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\$$\\frac{df}{dx}=\\var{a1}u^{\\simplify{{a1}-1}}.\\simplify{{a2}*{a3}x^{{a3}-1}}\$$

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\$$\\frac{df}{dx}=\\simplify{{a1}*{a2}*{a3}x^{{a3}-1}}({\\var{a2}x^{\\var{a3}}+\\var{a4}})^{\\simplify{{a1}-1}}\$$

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

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Differentiate the function:

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\$$f(x)=({\\var{a2}x^{\\var{a3}}+\\var{a4}})^\\var{a1}\$$

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Chain rule

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