How can I differentiate this expression?

[cris(c)]

Homework Statement

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

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.

 

[SammyS]

Homework Statement

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(f_2(\xi))G(f_3(\xi))d\xi

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.

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.

 

[cris(c)]

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.

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