# Homework Help: Compute Higher Order Mixed Derivative.

1. Jan 9, 2013

### smerhej

1. The problem statement, all variables and given/known data

Let f(u,v) be an infinitely differentiable function of two variables, and let g(x,y) = (x^2 + y^4, xy). If f$_{v}$ (5,2) = 1, f$_{uu}$ (5,2) = 2, f$_{vv}$ (5,2) = -2 and f$_{uv}$ (5,2) = 1, computer d^2(f o g)/dxdy at (2,1)

2. Relevant equations

The Chain Rule

3. The attempt at a solution

I set u = x^2 + y^4 and v = xy . Differentiating df/dx gives 2x(df/du) + y(df/dv) .[ df/dy gives 4y^3(df/du) + x(df/dv) but I don't think that's relevant ]. My idea from there was to use the results stated in the question to fill in these unknown derivatives.

Thanks!

2. Jan 9, 2013

### HallsofIvy

Yes, $\partial f/\partial x= 2x \partial f/\partial u+ y\partial f/\partial v$.

Now, differentiate that with respect to y:
$2x\partial^2 f/\partial u^2(4y^3)+ ...$

Last edited by a moderator: Jan 9, 2013
3. Jan 9, 2013

### lurflurf

Use (derive?) Faà di Bruno's formula.
Just include all possibilities
or do
chain rule
product rule (twice)
chain rule (twice)

$$(f \circ g)_{xy}=(f_u u_x+f_v v_x)_y \\ = f_u u_{xy}+f_v v_{xy}+(f_u)_y u_x+(f_v)_y v_x\\ =f_u u_{xy}+f_v v_{x y}+f_{uu} u_x u_y+f_{uv} u_x v_y+f_{vu} v_x u_y+f_{vv} v_x v_y$$
where subscripts denote differentiation and g(x,y)=(u,v)

4. Jan 10, 2013

### smerhej

Ok so I'm pretty sure I got the derivative worked out. First we're trying to figure out d(f o g)/dxdy. So differentiate with respect to y first then x. Doing so gives:

[ df/dy = 4y3 df/du + xdf/dv ]

Differentiating with respect to x gives:

[ 4y3d/dx(df/du) + df/dv +d/dx(df/dv) ]

Then, differentiating d/dx(df/du) and d/dx(df/dv), and putting those back into the equation above gives:

4y3d2f/du22x + d2f/dudvy +df/dv + d2f/dvdu2x + d2f/dv2y

If anyone wants to link me a "how to write math properly on this forum guide" that would be swell..

5. Jan 10, 2013