Is it possible to know that $\frac{\partial}{\partial y} \int_{-\infty}^{g(y)} f(x) dx = (g(y))\cdot \frac{\partial}{\partial y}g(y)$?

Question

Is it possible to know that $\frac{\partial}{\partial y} \int_{-\infty}^{g(y)} f(x) dx = f(g(y))\cdot \frac{\partial}{\partial y}g(y)$ without computing the $- \infty$ part explicitly?

Answer 1

I presume that we are looking at $f$ continuous?

$f$ does not have to go to zero at $-\infty$ but you need to know that $$ F(t)= \lim_{R\rightarrow +\infty} \int_{-R}^t f(x)\; dx $$ exists for some $t$. It is then automatically $C^1$ in all $t\in {\Bbb R}$. And you are simply calculating the derivative of the composed function $F(g(y))$.