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I am confused about this problem of finding the derivative of $e^y$ when differentiating with respect to $x$. The whole problem is to differentiate $y = x \, e^y$ with respect to $x$ but I get stuck on $\frac{d}{dx}(e^y)$.

I use the chain rule and end up with $(e^y)(y)(\frac{dy}{dx})$, derivative of the outside times inside times derivative of the inside, but when I look up online to check my answer it seems that $\frac{d}{dx}(e^y) = (e^y)(\frac{dy}{dx})$. I'm confused where my extra $y$ went?

Any help would be greatly appreciated.

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    $\begingroup$ Can you elaborate why you think there is an extra $y$ factor? The answer comes from $$\frac{de^y}{dx} = \frac{de^y}{dy}\cdot \frac{dy}{dx} = e^y\frac{dy}{dx}$$ $\endgroup$
    – peterwhy
    Commented May 15, 2016 at 21:28
  • $\begingroup$ I don't get the question: are you trying to find $\frac{dy}{dx}$ (or $\frac{d}{dx}e^y$) starting from the equation $y=xe^y$? $\endgroup$
    – yellon
    Commented May 15, 2016 at 21:29
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    $\begingroup$ He read $\frac d{dx}f(g(x))=f'(g(x))·g'(x)$ wrongly as $f'(?)·g(x)·g'(x)$, "derivative of the outside times inside times derivative of the inside" instead of "derivative of the outside [function] at the inside [function value] times derivative of the inside [function]". $\endgroup$
    – Lutz Lehmann
    Commented May 16, 2016 at 9:58

2 Answers 2

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$$\frac{\ d}{\ dx}e^y$$

First take the derivative like you "normally would":

$$e^y$$

Then take the derivative of the stuff substituted "inside", the stuff where an $x$ would usually be:

$$\frac{\ d}{\ dx} y=\frac{\ dy}{\ dx}$$

Multiply them together.

$$e^y • \frac{dy}{dx}$$

Summarized mathemetically,

$$\frac{du}{dx}=\frac{du}{dy}•\frac{dy}{dx}$$

Where here $u=e^y$.

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You can see it more clearly if you write it down this way:

$$\frac{d}{dx}e^{y}=\frac{d}{dx}e^{y(x)} = (e^{y(x)})'$$

So just apply the chain rule:

$$(e^{y(x)})' = y'(x)e^{y(x)} = \frac{dy}{dx}e^y$$

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  • $\begingroup$ Thank you! This helped also $\endgroup$
    – Paul zimmer
    Commented May 16, 2016 at 1:00
  • $\begingroup$ @Paulzimmer instead of commenting, you should rather upvote :) $\endgroup$
    – adjan
    Commented May 16, 2016 at 2:04

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