Compute Partials of F(z,w) Using Chain Rule

[hoffmann]

Use the chain rule to compute the partials of
F(z,w) = f(g_1(z,w),g_2(z,w),z,w)
where f(x,y,z,w)=x^2 +y^2 +z^2 −w^2
and g_1(z,w) = wcosz , g_2(z,w) = wsinz

Evaluate the partials at z = 0, w = 1. Confirm your result by writing out F explicitly as a function of z and w, computing its partial derivatives, and then evaluating at the same point.

I'm a little confused by the differentiation here:

so x = wcosz, y = wsinz thus:
x^2 = w^2cos^z
y^2 = w^2sin^2z
z = z^2
w = -w^2

what next?

 

[kurt.physics]

Answer: The chain rule states that the partial derivatives of F(z,w) with respect to z and w are given by ∂F/∂z = ∂f/∂g1 · ∂g1/∂z + ∂f/∂g2 · ∂g2/∂z + ∂f/∂z and ∂F/∂w = ∂f/∂g1 · ∂g1/∂w + ∂f/∂g2 · ∂g2/∂w + ∂f/∂w. Using the given functions for f and g1 and g2, we can calculate the partials of F(z,w):∂F/∂z = 2z − wsinz∂F/∂w = 2w − wcosz Evaluating at z = 0 and w = 1, we get ∂F/∂z = 0∂F/∂w = 2To confirm our result, we can write out F explicitly as a function of z and w and compute its partial derivatives: F(z,w) = (wcosz)^2 + (wsinz)^2 + z^2 - w^2 ∂F/∂z = 2z − wsinz∂F/∂w = 2w − wcosz Evaluating at z = 0 and w = 1, we again get ∂F/∂z = 0∂F/∂w = 2