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.