Partial derivation of $u:\mathbb{R^2} \to \mathbb{R} \text{ with } (x,y) \to \int_{0}^{xy}\log\left(v(t)\right) dt$

Let $$u:\mathbb{R^2} \to \mathbb{R} \text{ with } (x,y) \to \int_{0}^{xy}\log\left(v(t)\right) dt$$ whereby $$v: \mathbb{R} \to \mathbb{R}$$ is a continuous function.

I want to find the set of points $$D$$, in which this function is partially differentiable and calculate its partial derivatives and gradient there.

So I have to look for a fixed $$y \in \mathbb{R}$$ in the function $$v(\cdot, y):\mathbb{R} \to \mathbb{R}; t \to \int_{0}^{yt}\log\left(v(s)\right) ds$$

I know that the fundamental theorem of calculus is that $$\frac{dg(\cdot, y)}{dt} (t) = yv(yt)$$ for all $$t \in \mathbb{R}$$.

So $$\frac{\partial g}{\partial x} (x,y) = yv(xy)$$

How do I proceed analogously to determine $$\frac{\partial g}{\partial y}$$

The function $$u$$ is only defined, if $$v(t)>0$$ for all $$t \in \mathbb R$$.

Define the function $$F$$ by $$F(s):=\int_{0}^{s}\log\left(v(t)\right) dt$$.

By the the fundamental theorem of calculus we have $$F'(s)= \log(v(s)$$.

The function $$u$$ is now given by $$u(x,y)=F(xy)$$, hence

$$u_x(x,y)=F'(xy)y= y \log(v(xy)$$

and

$$u_y(x,y)=F'(xy)x= x \log(v(xy)$$.