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How to obtain $y'$ from $e^{y}=x^{\ln x}$?

This is what I did: $$\ln e^y = \ln x^{\ln x}$$ $$y = \ln ^2 x$$ $$y' = \frac{2 \ln x}{x}$$ Is this correct? When I compare it with an online derivative calculator, the result they gave was different, and they used implicit differentiation instead.

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    $\begingroup$ Seems fine to me. Graphs look good too. $\endgroup$ – mvw Feb 16 '17 at 9:11
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    $\begingroup$ I'd avoid $\ln^2x$, which is ambiguous, preferring $(\ln x)^2$; this said, you did perfectly right. $\endgroup$ – egreg Feb 16 '17 at 9:43
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So if we use implicit differentiation:

$$e^{y}\frac{\mathrm{d}y}{\mathrm{d}x}=2\ln(x)x^{\ln(x)-1}$$

But $e^{y} = x^{\ln(x)}$, so:

$$\frac{\mathrm{d}y}{\mathrm{d}x}=2\ln(x) \frac{x^{\ln(x)-1}}{x^{\ln(x)}}=\frac{2\ln(x)}{x}$$

So you are quite right in your answer (and your approach is absolutely fine too!)

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$$(e^y)'=e^yy'$$ and $$(e^{ln^2x})'=(\ln^2x)'e^{ln^2x}=2\frac{\ln x}xe^{ln^2x}.$$

Then dividing by $e^y=e^{ln^2x}$,

$$y'=2\frac{\ln x}x.$$

As expected, same answer, though a little more tedious.


This is $$g(y)=f(x),\\ g'(y)y'=f'(x),\\ y'=\frac{f'(x)}{g'(y)}=\frac{f'(x)}{g'(g^{-1}(f(x)))}$$ versus $$y=g^{-1}(f(x)),\\ y'=\frac{f'(x)}{g'(g^{-1}(f(x)))} $$ by the formula for the inverse function and the chain rule.

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