Second Partial Derivative of Compound Function of two variables

Question

Given the function to differentiate: $f\left ( u(x,y),v(x,y) \right )$

Where,

$\left\{\begin{matrix} u=xe^{-y}\\ v=y \end{matrix}\right.$

How do I calculate $\frac{\partial ^{2}f}{ \partial x \partial y}$ in terms of the same notation?

Does anyone have any idea?

Answer 1

We know that:

$\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial x}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial x}$

$\dfrac{\partial f}{\partial y}=\dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial y}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial y}$

So, you can now rewrite the main problem like below:

$\dfrac{\partial^{2}f}{ \partial x \partial y}=\dfrac{\partial}{\partial x}(\dfrac{\partial f}{\partial y})=\dfrac{\partial}{x}(\dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial y}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial y})$

$=\dfrac{\partial}{\partial u}(\dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial y}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial y}) . \dfrac{\partial u}{\partial x}+\dfrac{\partial}{\partial v}(\dfrac{\partial f}{\partial u} \dfrac{\partial u}{\partial y}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial y}) . \dfrac{\partial v}{\partial x}$