0
$\begingroup$

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 $$\frac{\partial g}{\partial y}$$

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy