Compute Higher Order Mixed Derivative.

[smerhej]

Homework Statement

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)

Homework Equations

The Chain Rule

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!

 

[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)+ ...

 

[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 \\<br /> = f_u u_{xy}+f_v v_{xy}+(f_u)_y u_x+(f_v)_y v_x\\<br /> =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)

 

[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..

 

[HallsofIvy]

https://www.physicsforums.com/showthread.php?p=3977517&posted=1#post3977517