1. ## Confused about partial derivatives of higher order...

After literally hours of "no, and what and how", I need help with the following task:

f(u,v)= g(x(u,v),y(u,v))

x(u,v)= u^2- v^2
y(u,v)= 2uv

Expressing f'u and f'v as partial derivatives of g I get:

f'u= (dg/dx)*(dx/du) + (dg/dy)*(dy/du)
f'v= (dg/dx)*(dx/dv) + (dg/dy)*(dy/dv)

(written in crooked d's)

Now I'm supposed to find all functions f that satisfy u*f'u - v*f'v=0

The hardest part (which confuse me the most because there are so many steps involving the chain rule is the following): "Express f''uu+ f''vv through partial derivatives of g. Simplify as much as possible"

3. Originally Posted by tinyone
After literally hours of "no, and what and how", I need help with the following task:

f(u,v)= g(x(u,v),y(u,v))

x(u,v)= u^2- v^2
y(u,v)= 2uv

Expressing f'u and f'v as partial derivatives of g I get:

f'u= (dg/dx)*(dx/du) + (dg/dy)*(dy/du)
f'v= (dg/dx)*(dx/dv) + (dg/dy)*(dy/dv)

(written in crooked d's)

Now I'm supposed to find all functions f that satisfy u*f'u - v*f'v=0
So you know that $f'_u = \frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u}$. You can write $\frac{\partial g}{\partial x}$ as $g'_x$, and $\frac{\partial g}{\partial y}$ as $g'_y$ (to save writing). You can also find $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial u}$ (from the formulas for x(u,v) and y(u,v)). That has the effect of writing $f'_u$ as

$f'_u = 2ug'_x + 2vg'_y\qquad(*)$.

Now do the same for the other partial derivative $f'_v$.

Having done that, multiply the formula for $f'_u$ by u, multiply the formula for $f'_v$ by v, and subtract. That will tell you something about the possible functions that satisfy $uf'_u - vf'_v = 0$.

Originally Posted by tinyone
The hardest part (which confuse me the most because there are so many steps involving the chain rule is the following): "Express f''uu+ f''vv through partial derivatives of g. Simplify as much as possible"
Differentiate the formula (*) partially with respect to u, using the chain rule and the product rule: $f''_{uu} = 2g'_x + 2u\frac{\partial}{\partial u}(g'_x) + 2v\frac{\partial}{\partial u}(g'_y)$. Now you need to figure out what $\frac{\partial}{\partial u}(g'_x)$ and $\frac{\partial}{\partial u}(g'_y)$ are. To do that, use the formula (*), but replace $f$ by $g'_x$. That tells you that $\frac{\partial}{\partial u}(g'_x) = 2ug''_{xx} + 2vg''_{xy}$ (and you can get a similar expression for $\frac{\partial}{\partial u}(g'_y)$).

That way, you get an expression for $f''_{uu}$ in terms of the second partial derivatives of g. Do the same for $f''_{vv}$, add, and simplify ... .