Suppose you are given that $f$ is a differentiable function of two variables $u$ and $v$. Let $g(r,s)=f(r^2-s^2, 2rs)$. Compute $\nabla g(r,s)$ in terms of $\partial f\over\partial u$, $\partial f\over\partial v$ and other functions.

I expressed $f$ as $f(u,v)=f(r^2-s^2, 2rs)$ and set $u=r^2-s^2$ and $v=2rs$. To calculate the gradient I did:

$$\nabla g(r,s)=\frac{\partial g}{\partial r}\mathbf i + \frac{\partial g}{\partial s}\mathbf j$$

From there:

$$=\left({\partial f\over\partial u}{\partial u\over\partial r}\right)\mathbf i + \left({\partial f\over\partial v}{\partial v\over\partial s}\right)\mathbf j$$

This is the step where I lost points but I can't seem to figure out where I went wrong.


In replacing $\frac{\partial g}{\partial r}$ with $\left({\partial f\over\partial u}{\partial u\over\partial r}\right)$, you are treating $f$ as if it were a function of $u$ alone, and similarly in replacing $\frac{\partial g}{\partial s}$ with $\left({\partial f\over\partial v}{\partial v\over\partial s}\right)$, you are treating $f$ as if it were a function of $v$ alone.

The correct formulas are:$$\frac{\partial g}{\partial r} = {\partial f\over\partial u}{\partial u\over\partial r} + {\partial f\over\partial v}{\partial v\over\partial r}\\\frac{\partial g}{\partial s} = {\partial f\over\partial u}{\partial u\over\partial s} + {\partial f\over\partial v}{\partial v\over\partial s}$$

| cite | improve this answer | |
  • $\begingroup$ If I understand correctly: $g(r,s)=f(r^2-s^2,2rs)=f(u,v)$ Set $u=r^2-s^2$ and $v=2rs$ Branch diagram: f-->(u or v), u-->(r or s), v-->(r or s) Is my logical approach correct? $\endgroup$ – Omrane Dec 5 '15 at 22:15
  • $\begingroup$ I have no clue what you mean by "Branch diagram: $f\ u\ v\ r\ s\ r\ s$". The rest is okay. But the problem is still that you overlooked the dependence of $f$ on $v$ when substituting for $\partial g\over\partial r$, and of $f$ on $u$ when substituting for $\partial g\over\partial s$ $\endgroup$ – Paul Sinclair Dec 5 '15 at 22:20
  • $\begingroup$ I fixed the Branch Diagram, didn't know how to implement in comment. But anyways it follows what you said so I think I got the heck of it all. Thanks again for your help. $\endgroup$ – Omrane Dec 5 '15 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.