Let $\Phi(\cdot)$ be the CDF of a standard normal random variable $Z \sim N(0,1)$. Define a function
$$ F(x) = \int_{-\ln (x)}^{x^2} \Phi(t+x) dt. $$
Question: what is $F'(1)$, where $F'(x)$ is the first derivative of $F$?
Now I have broken the integral in two pieces as follows
$$ F(x) = -\int_{0}^{\ln x} \Phi(t+x) dt + \int_{0}^{x^2} \Phi(t+x) dt $$
and want to differentiate both integrals according with the formula
$$ G(x) = \int_0^{\phi(x)} h(x,y) dy \Rightarrow G'(x) = \int_0^{\phi(x)} h_x(x,y) dy +h(x, \phi(x))\phi(x)'. $$
However, at this point I am not sure how to appy this formula. Indeed, if I want to use it to determine for example the first integral
$$ (-\int_{0}^{\ln x} \Phi(t+x) dt)' $$
I am forced to compute
$$ \int_{0}^{\ln x} \frac{\partial}{\partial x}\Phi(t+x) dt $$
and I am stuck here. How is this integral to be done?