# How can I differentiate this expression?

1. Jan 21, 2012

### cris(c)

1. The problem statement, all variables and given/known data
Consider three univariate distinct functions $f_1(x),f_2(y),f_3(y)$. Let H be given by the following integral:

$H=\int_{0}^{f_1(x)} G(f2(\xi))G(f3(\xi))d\xi$

3. The attempt at a solution
Then, computing dH/dy should give zero. However, I am not certain of this because the chain rule would give me:

$\frac{dH}{dy}=\frac{\partial H}{\partial y} + \frac{\partial H}{\partial f_2}\frac{\partial f_2}{\partial y}+ \frac{\partial H}{\partial f_3}\frac{\partial f_3}{\partial y}$

and $\frac{\partial H}{\partial f_j}\frac{\partial f_j}{\partial y}$ are nonzero.

2. Jan 21, 2012

### SammyS

Staff Emeritus
I had hoped one of our "experts" such as Dick, or micromass, or ... would have responded to this, but after looking at it a few times, I'll take a stab at it. (An informed stab at that.)

The integrand $\displaystyle G(f_2(\xi))G(f_3(\xi))$ is a function of ξ. You are integrating over ξ and the limits of integration are 0 and f1(x). Therefore, H is a function of x and only of x, so of course, $\displaystyle \frac{dH}{dy}=0\,.$

If you intended the limits of integration to be 0 and f1(y), then finding $\displaystyle \frac{dH}{dy}$ makes sense.

If $\displaystyle H=\int_{0}^{f_1(y)} G(f_2(\xi))G(f_3(\xi))d\xi\,,$ the H(y) is the anti-derivative of $\displaystyle G(f_2(\xi))G(f_3(\xi))$ evaluated at ξ=f1(y), minus a constant.

The derivative of this result (w.r.t. y) should be pretty obvious, if this is indeed the question.

3. Jan 22, 2012

### cris(c)

Thanks SammyS. I thought exactly the same as you. However, I still don't see why the chain rule would be invalidated in this case. I know that so long as y does not appear in the limits of integration the integral should not change with y, but why the chain rule doesn't appear to say the sameÉ