- #1

- 182

- 0

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

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.

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.

Last edited: