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

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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)=\\var{a1}ln(u)\$$

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

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

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

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\$$\\frac{df}{dx}=\$$ [[0]]

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Differentiate the function:

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

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Chain rule

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