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.

Let $x^2-t^2=m$.
And  $-2tdt=dm$.
$\int_0^x tf(x^2-t^2) dt=\int_{x^2}^0 tf(m) \frac{dm}{-2t}$
$=-\frac{1}{2} \int_{x^2}^0 f(m) dm$
Hence,
$\frac{d}{dx}\int_0^xtf(x^2-t^2) dt=-\frac{1}{2} \frac{d}{dx}  \int_{x^2}^0 f(m) dm=xf(x^2)$
