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\\(f(x)=(\\var{a1}x^\\var{a2}+\\var{a3})e^{\\var{a4}x+\\var{a5}}\\)

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Recall the product rule if \\(f(x)=u.v\\) where \\(u\\) and \\(v\\) are both functions of \\(x\\) then

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\\(\\frac{df}{dx}=v.\\frac{du}{dx}+u.\\frac{dv}{dx}\\)

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let \\(u=\\var{a1}x^\\var{a2}+\\var{a3}\\)  and  \\(v=e^{\\var{a4}x+\\var{a5}}\\)

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\\(\\frac{du}{dx}=\\var{a2}*\\var{a1}x^{\\var{a2}-1}\\)  and  \\(\\frac{dv}{dx}=\\var{a4}*e^{\\var{a4}x+\\var{a5}}\\)

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\\(\\frac{df}{dx}=e^{\\var{a4}x+\\var{a5}}*\\var{a2}*\\var{a1}x^{\\var{a2}-1}+(\\var{a1}x^\\var{a2}+\\var{a3})*\\var{a4}*e^{\\var{a4}x+\\var{a5}}\\)

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\\(\\frac{df}{dx}=\\simplify{e^({a4}x+{a5})*{a1}*{a2}x^{{a2}-1}+({a1}x^{a2}+{a3})*{a4}*e^({a4}x+{a5})}\\)

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\\(\\frac{df}{dx}=(\\simplify{{a1}*{a4}x^{a2}+{a1}*{a2}x^{{a2}-1}+{a3}*{a4}})\\simplify{e^({a4}x+{a5})}\\)

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\\(\\frac{df}{dx}=\\)[[0]]

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Differentiate the function

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\\(f(x)=(\\var{a1}x^\\var{a2}+\\var{a3})e^{\\var{a4}x+\\var{a5}}\\)

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Product rule

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