Partial derivatives of $(x-y)^x$

Question

Let $h:$ {$(x,y) \in \mathbb{R}^2: x > y > 0$} $ \to \mathbb{R}$ with $$h(x,y) := (x-y)^x$$

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 would write:

$D(h)=\{(x,y) \mid x,y \in \mathbb{R^+ \text{\ \{0}}\}$}

$$\frac{\partial h}{\partial x} = (x-y)^x$$

and

$$\frac{\partial h}{\partial y} = (x-y)^x$$

But what is the partial derivative of $(x-y)^x$ for x? for y?

Answer 1

\begin{align} h(x,y) &=(x-y)^x\\ \ln h &=\ln (x-y)^x =x\ln (x-y)\\ \frac{\partial(\ln h)}{\partial x} =\frac{\left(\frac{\partial h}{\partial x}\right)}{h} &= \ln (x-y) +\frac{x}{x-y}\\ \frac{\partial h}{\partial x} &=h\cdot \frac{\partial(\ln h)}{\partial x} =(x-y)^x \ln(x-y) +x(x-y)^{x-1}\\ \frac{\partial h}{\partial y} &=x(x-y)^{x-1} \cdot \frac{\partial(x-y)}{\partial y} =-x(x-y)^{x-1} \end{align}

Answer 2

We have $D(h)=\{(x,y) \in \mathbb{R}^2: x > y > 0\}$.

Hint for the derivatives: $h(x,y)=(x-y)^x= e^{x \ln (x-y)}$.