Why isn't the derivative of $e^{-x}$ equal to $-e^{-2x}$?

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.

• It looks like you forgot the $u'$ factor. – 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

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

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

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