How Do You Compute Derivatives in Spherical Coordinates Using the Chain Rule?

[Tony11235]

Let f: \Re^3 \rightarrow \Re be differentiable. Making the substitution

x = \rho \cos{\theta} \sin{\phi}, y = \rho \sin{\theta} \sin{\phi}, z = \rho \cos{\phi}

(spherical coordinates) into f(x,y,z), compute (partially) df/d(rho), df/d(theta), and df/d(phi) in terms of df/dx, df/dy, and df/dz.

I'm just not sure I understand the question. Does it involve pulling out a very long chain rule?

 

[lurflurf]

Let f: \Re^3 \rightarrow \Re be differentiable. Making the substitution

x = \rho \cos{\theta} \sin{\phi}, y = \rho \sin{\theta} \sin{\phi}, z = \rho \cos{\phi}

(spherical coordinates) into f(x,y,z), compute (partially) df/d(rho), df/d(theta), and df/d(phi) in terms of df/dx, df/dy, and df/dz.

I'm just not sure I understand the question. Does it involve pulling out a very long chain rule?

It involves the chain rule, not sure what you mean about the very long part.
\frac{\partial f}{\partial\rho}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\rho}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\rho}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\rho}
\frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\theta}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\theta}
\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial x} \ \frac{\partial x}{\partial\phi}+\frac{\partial f}{\partial y} \ \frac{\partial y}{\partial\phi}+\frac{\partial f}{\partial z} \ \frac{\partial z}{\partial\phi}
The general form of the chain rule being
\frac{\partial f}{\partial x}=\sum_{k=1}^n \frac{\partial f}{\partial u_k} \ \frac{\partial u_k}{\partial x}
where
f=f(u_1(x),u_2(x),...,u_{n-1}(x),u_n(x))