Finding the derivative of an integral?

Question

I'm reading Bliss's "Calculus of Variations," where he gives the function

I(a) = ∫ f(y' + aη') dx

where y and η are continuous and piecewise differentiable functions of x, y'=dy/dx and η'=dη/dx.  Limits of integration are x_1 and x_2.

The task is to differentiate I(a) with respect to a then set a=0.  Bliss says the result is "readily seen to be"

I'(0) = ∫ f_(y') η' dx

where f_(y') is the derivative of f(y') with respect to y'.  Limits of integration are again x_1 and x_2.  

I know I'm going to kick myself when I see the answer, but it's been several years since Multivariable Calculus and I'm just not seeing it on my own.

Ryoma · Accepted Answer

Hi

I'll assume that a is independent of x. In that case, we can apply Leibnitz rule of differentiation under the integral sign (it says that we can move the differential operator over the integral sign) along with the chain rule (since f(y' + aη') is a composite function) to obtain:
I'(a) = d/da ∫ f(y' + aη') dx
= ∫ d/da f(y' + aη') dx
= ∫ d/d(y' + aη') f(y' + aη') * d/da (y' + aη') dx
= ∫ f'(y' + aη') * η' dx

Plugging in a = 0 yields:
I'(0) = ∫ f'(y' + 0*η') * η' dx
= ∫ f'(y') * η' dx

I hope this helps!

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Finding the derivative of an integral?

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