I have the integral from $-\infty$ to $y^2$ of the function $(e^{-|x|})$ and I need to find the derivative of this. That is,

$$\frac{d}{dy} \int_{-\infty}^{y^2} e^{-|x|}\,dx$$

Usually derivative of integral is just the function, but I'm not sure in this case. Should I set up limits or how else should I approach this?

I thought that the derivative of an integral from $-\infty$ to $y^2$ of $(e^{-|x|}dx)$ would be $e^{-|x|}$

  • $\begingroup$ You can define $f(t) = \int_{-\infty}^{t} e^{-|x|}dx$. Then the function you are interested in is $g(y) = f(y^2)$. So you differentiate with the chain rule. $\endgroup$
    – Michael
    Sep 14, 2014 at 18:37
  • $\begingroup$ also note that b/c of the absolute value you need to split the integral. $\endgroup$
    – Varun Iyer
    Sep 14, 2014 at 18:39
  • $\begingroup$ @VarunIyer : You could split the integral if you first wanted to solve for the exact integral. But this problem does not require you to do any integration. $\endgroup$
    – Michael
    Sep 14, 2014 at 18:41
  • 1
    $\begingroup$ If he would like to know how to solve the integral, its another solution :). $\endgroup$
    – Varun Iyer
    Sep 14, 2014 at 18:42
  • $\begingroup$ so lim as a->-infinity of integral from a to 0 e^-|y^2| + integral from 0 to t e^-|y^2| ? $\endgroup$
    – user176017
    Sep 14, 2014 at 18:42

2 Answers 2


Let $$F(u)=\int_{-\infty}^u e^{-|x|}\,dx.$$ We want $\frac{d}{dy}(F(y^2))$. Let $u=y^2$. By the Chain Rule, we have $$\frac{d}{dy}(F(y^2))=2y\frac{d}{du} F(u).$$

By the Fundamental Theorem of Calculus, we have $\frac{d}{du}F(u)=e^{-|u|}$.

Remark: If the $-\infty$ part makes you uncomfortable, note that $F(y^2)=\int_{-\infty}^0 e^{-|x|}\,dx+\int_0^{y^2} e^{-|x|}\,dx$, and the first integral is just a constant.


This is another possible solution

Much more lengthier, but if you would like to know how to derive:

So the integral is:

$$\int_{-\infty}^{y^2} e^{-|x|} dx = \int_{-\infty}^{0} e^{x} dx + \int_{0}^{y^2} e^{-x} dx$$

In this case, it would be easier to find the integral then perform the derivative:

So for the first we have that:

$$\int_{-\infty}^{0} e^{x} dx = \lim_{b\to-\infty}\int_{b}^{0} e^{x} dx \lim_{b\to-\infty}\left[e^x\right]_{b}^{0} = \lim_{b\to-\infty} 1 - e^b = 1- 0 = 1$$

For the second one:

$$\int_{0}^{y^2} e^{-x} dx = \left[-e^{-x}\right]_{0}^{y^2} = -e^{-y^2} - (-e^{0}) = -e^{-y^2} + 1$$

So our final answer is:

$$-e^{-y^2} + 1 + 1 = -e^{-y^2} + 2$$

Taking the derivative, we get:

$$\left(-e^{-y^2}\right)' = 2y*e^{-y^2}$$

Comment if you have questions.

  • $\begingroup$ The first of the two integrals is constant; no need to compute it at all! $\endgroup$
    – user14972
    Sep 14, 2014 at 18:50
  • $\begingroup$ @Hurkyl I mentioned that in my comments above. This is another way to do the calculations. More lengthier, but if you like deriving, then have fun :). $\endgroup$
    – Varun Iyer
    Sep 14, 2014 at 18:52

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