What is the derivative of $y = \int_{0}^x t f(x^2-t^2)dt$? Let $y = \int_{0}^x t f(x^2-t^2)dt$, where $f$ is differentiable. How to compute $\frac{dy}{dx}$?
My solution: I think that 
$$
\frac{dy}{dx} = \int_{0}^x 2x t f'(x^2 - t^2)dt + tf(0). 
$$
Is this correct? Thank you very much.

If $g:\Bbb R^2 \rightarrow \Bbb R$, $\Delta:\Bbb R \rightarrow \Bbb R^2$ is the diagonal map, and $f(x)=g(x,x)$ (or $f=g\circ \Delta$) then the chain rule gives $f'(x)=g_1(x,x)+g_2(x,x)$ (where $g_i$ is the $i$th partial derivative).
Letting $g(x,y)=\int_{0}^x t f(y^2-t^2)dt$ gives the desired result.
