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Hey so I was wondering how to differentiate $(e^2)^x$ without using the chain rule.

I tried but I always end up using the chain rule in this case.

Would appreciate some help! (No hints please).

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    $\begingroup$ I'm confused. You want help, but you don't want hints? $\endgroup$ – Y. Forman Jun 12 '18 at 14:06
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    $\begingroup$ Hint: Consider its Maclaurin series. $\endgroup$ – poyea Jun 12 '18 at 14:07
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    $\begingroup$ Use the limit definition and properties of $e$. Sorry if this is a hint. $\endgroup$ – Randall Jun 12 '18 at 14:08
  • $\begingroup$ Forman, if you are confused as to why someone needs help but doesn’t want hint answers, then maybe its not for you. $\endgroup$ – user549904 Jun 12 '18 at 14:17
  • $\begingroup$ user549904, if you think other people should do your work and you cannot deal with just hints, maybe math is not for you. $\endgroup$ – Torsten Schoeneberg Jun 12 '18 at 14:20
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Indeed, using the chain rule is certainly the easiest: $$f(x)=e^{2x}\to f'(x)=2e^{2x}$$


You could however, use the product rule: $$f(x)=(e^2)^x=e^xe^x$$ Then $p=q=e^x$, and $p'=q'=e^x$

Product rule is $f'(x)=p'q+q'p\to e^xe^x+e^xe^x=2e^{2x}$

Or the quotient rule can also be used: $$f(x)=(e^2)^x=\frac{e^x}{e^{-x}}$$ Then $p=p'=e^x$, $q=e^{-x}\to q'=-e^{-x}$

Quotient rule is $f'(x)=\frac{p'q-q'p}{q^2}\to\frac{e^xe^{-x}-e^x(-e^{-x})}{e^{-2x}}=\frac{2}{e^{-2x}}=2e^{2x}$

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If you know the rule

$$(a^x)'=\ln a\,a^x$$ then

$$((e^2)^x)'=\ln e^2\,(e^2)^x=2(e^2)^x.$$


Alternatively,

$$(e^{2x})'=\lim_{h\to0}\frac{e^{2(x+h)}-e^{2x}}h=e^{2x}\lim_{h\to0}\frac{e^{2h}-1}h=2e^x\lim_{2h\to0}\frac{e^{2h}-1}{2h}=2e^{2x}.$$

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