Chain Rule, second partials.

[STEMucator]

Homework Statement

I'm curious to know if I'm actually doing this correctly.

Suppose f(x,y) is a function where x = p(s,t) and y = g(s,t) so that w(s,t) = f(x,y).

Compute w_s and then w_st

Homework Equations

Chain Rule.

The Attempt at a Solution

So! Let's compute w_s first. Whenever I use a subscript I refer to the partial with respect to that variable.

$w_s = f_x x_s + f_y y_s$

That was nice and easy... now for the hard part that I'm not sure of. Let's compute w_st :

$w_{st} = (f_{xx} x_t + f_{xy} y_t)x_s + f_x x_{st} + (f_{xy} x_t + f_{yy} y_t)y_s + f_y y_{st}$

I think this is correct, but with all the variables floating around I'm not entirely sure I didn't miss anything. I would appreciate it very much if someone could verify this for me as I don't want to have a 'blah' moment when my exam happens.

 

[mighty2000]

Hi there! It looks like you are on the right track with your solution. The chain rule is definitely the correct method to use in this situation.

For your calculation of wst, you have correctly included the terms involving the second-order partial derivatives (fxx, fxy, fyy). However, you are missing one term - the one involving the first-order partial derivatives of x and y. The correct expression for wst would be:

w_{st} = (f_{xx} x_t + f_{xy} y_t)x_s + f_x x_{st} + (f_{xy} x_t + f_{yy} y_t)y_s + f_y y_{st} + (f_{x} x_s + f_{y} y_s) t_s

Note that the last term in parentheses involves the first-order partial derivatives of x and y with respect to s, multiplied by the partial derivative of t with respect to s. This term is necessary because both x and y are functions of s, so their partial derivatives with respect to t also depend on s.

I hope this helps clarify your solution and good luck on your exam!