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I am reading a book on analysis in which using generalized mean value theorem for $g(x)=e^xf(x)$. it is shown that $\displaystyle \frac{g'(c)}{e^c}=\frac{g(x)-g(a)}{e^x-e^a}$ for some $c\in (a,x)$. But I am unable to show how they got it. Could you please show me how this step came?

Edit: Another way using generalized mean value theorem $\displaystyle \frac{e^xf(x)-e^af(a)}{e^x-e^a}={f(c)+f'(c)}$. But I am finding it difficult to get this step.

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I think you do not need the special form of g to prove this equality, as it is a direct application of the generalized mean value theorem for an arbitrary function g and exp.

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  • $\begingroup$ Okay I got it . Thanks for your help. $\endgroup$ – Ziya Feb 26 '18 at 17:19

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