Differentiating an Integral: Proving the Chain Rule for Derivatives

[mathboy]

Notation: I^x means the integral sign from 0 to x

Question: Let f be a smooth function. Prove that
d/dx [I^x f(x,y)dy] = f(x,x) + I^x (d/dx)f(x,y)dy
using the chain rule for derivatives (do NOT use Leibnitz's rule for differentiating an integral).

I don't know how to express I^x f(x,y)dy as a composition of two functions. I've tried defining F(x,z) = I^x f(z,y)dy and then compute d/dx[F(x,x)] but I can't get anywhere. I do know that by the fundamental theorem of calculus that
d/dy [I^x f(x,y)dy] = f(x,y)
but I can't seem to incorporate it here.

Please someone tell me what composition I'm supposed to take the derivative of.

 

[morphism]

What are the conditions on f? For instance the right hand side won't make sense if the first partial (i.e. wrt x) of f isn't continuous almost everywhere. Even assuming this, I don't see how we can proceed without using Leibniz, or at least implicitly using the proof of Leibniz. How else are we going to get the partial under the integral sign?

See this thread for more information. Most of the methods there can be suitably adapted to solve your problem.

 

[mathboy]

f is a smooth function. And the question can be done using the chain rule according to the question (the question comes from a section devoted to the chain rule), and the fundamental theorem of calculus will have to be used.

I'm looking over the thread you pointed out and some chain rules are being used there.

 

[morphism]

Yup - but notice that the basic version of Leibniz is also used. I suppose you can also incorporate its proof (see homology's post at the end of the first page) into your solution.

 

[mathboy]

Ok, I got it now. The set up is:

h(x) = (z(x),x)
F(h(x)) = I^z(x) f(x,y)dy

Then the chain rule gives the answer exactly upon using z(x) = x. Thanks.