Question

Consider the function g defined by:

g(x,y) = cos (xπ $√y$ ) + 1/log₃(x - y)

Determine ∂²g/∂y∂x

Answer 1

Function $g(x,y)$ has the form: $g(x,y)=\cos(x\pi\sqrt{y})+\frac{1}{log_3(x-y)}=\cos(x\pi\sqrt{y})+\frac{\ln(3)}{ln(x-y)}$.

We receive:

$\frac{\partial g}{\partial y}=-\frac{x\pi}{2\sqrt{y}}\sin(x\pi\sqrt{y})+\frac{\ln(3)}{(\ln(x-y))^2}\cdot\frac{1}{x-y}\\$. $\,\frac{\partial^2g}{\partial y\partial x}=-\frac{\pi}{2\sqrt{y}}\sin(x\pi\sqrt{y})-\frac{x\pi^2}{2}\cos(x\pi\sqrt{y})-\frac{(\ln(3))(2+\ln(x-y))}{(\ln(x-y))^3}\cdot\frac{1}{(x-y)^2}$.

Answer: $\frac{\partial^2g}{\partial y\partial x}=-\frac{\pi}{2\sqrt{y}}\sin(x\pi\sqrt{y})-\frac{x\pi^2}{2}\cos(x\pi\sqrt{y})-\frac{(\ln(3))(2+\ln(x-y))}{(\ln(x-y))^3}\cdot\frac{1}{(x-y)^2}$