Question

Find all the second partial derivatives. f(x, y) = sin2(mx + ny)

Answer 1

We have:
\[f(x,y)=sin[2(mx+ny)]\]
For a function \(f(x,y)\), there exist four possible partial derivatives:
\[\frac{\partial }{\partial x}\left[\frac{\partial f}{\partial x}\right]=f_{xx}=a\]
\[\frac{\partial }{\partial x}\left[\frac{\partial f}{\partial y}\right]=f_{xy}=b\]
\[\frac{\partial }{\partial y}\left[\frac{\partial f}{\partial x}\right]=f_{yx}=c\]
\[\frac{\partial }{\partial y}\left[\frac{\partial f}{\partial y}\right]=f_{yy}=d\]
We will calculate each!

Answer 2

Good so far? Makes sense?

Answer 3

Lets start:
\[a=\frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right]=\frac{\partial}{\partial x}\left[\frac{\partial\bigg(\sin[2(mx+ny)]\bigg)}{\partial x}\right]=\frac{\partial}{\partial x}\left[\frac{\partial\Big(\sin(2mx+2ny)\Big)}{\partial x}\right]\]
\[=\frac{\partial}{\partial x}\left[2m\cos(2mx+2ny)\right]=-4m^2\sin(2mx+2ny)\]

Answer 4

Cool?