I'm sorry if this is a stupid question:

If you have an equation $y = (e^x)^{-1}$, why is it that when I use chain rule with a substitution of $u=e^x$ (giving $y = u^{-1}$), I get the wrong derivative, namely $-e^{-2x}$?

I'm struggling to find an intuition.

  • 7
    $\begingroup$ It looks like you forgot the $u'$ factor. $\endgroup$ – J.G. Dec 17 '18 at 23:03

We need to use by chain rule and by $u=e^x$ we have

$$y=\frac 1u \implies y'=-\frac1{u^2}\cdot \color{red}{u'}=-\frac1{e^{2x}}\cdot e^x=-\frac1{e^{x}}$$

which agrees with the direct evaluation $y=e^{f(x)} \implies y'=f'(x)e^{f(x)} $ that is

$$y=e^{-x} \implies y'=-e^{-x}$$

  • $\begingroup$ Thank you very much! Wow that was a very silly oversight on my side. $\endgroup$ – Devansh Shah Dec 17 '18 at 23:40
  • $\begingroup$ You’re welcome! We always learn from mistakes, next time you’ll be aware about that! Bye $\endgroup$ – gimusi Dec 17 '18 at 23:41
  • $\begingroup$ Hi Gimusi. Why where you suspended this time? $\endgroup$ – user370967 Jan 15 at 17:06
  • $\begingroup$ @gimusi How come you are not active anymore? $\endgroup$ – Maria Mazur Feb 26 at 23:10
  • $\begingroup$ Probably because of all the suspensions. $\endgroup$ – user370967 Mar 3 at 15:56

Looks like you are using a combination of the power rule and the chain rule.

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

You could also do this with just the chain rule:

$y = e^{-x}$

let $u = -x$

$y = e^u\\ \frac {dy}{dx} = \frac {dy}{du}\frac {du}{dx} = (e^u)(-1) = -e^{-x}$


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