1
$\begingroup$

Ok guys, I'm given this smooth function $\varphi(u,v)$ defined in $R^2$. So that $f(x,y)=\varphi(xy,\frac{x}{y})$. I have to find all partial derivatives of second order of $f$ using the partial derivatives of $\varphi$. I know how to find the partial derivative of "normal" functions like $\frac{xy}{x+y}$ or something like that, but this kind of problem I have no idea how to do. Any ideas and solutions are welcomed $\ddot \smile $

$\endgroup$
2
$\begingroup$

Use the chain rule: $$ \frac{\partial}{\partial x} \phi(u(x,y),v(x,y)) = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v} \frac{\partial v}{\partial x}, $$

$$ \frac{\partial}{\partial y} \phi(u(x,y),v(x,y)) = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial \phi}{\partial v} \frac{\partial v}{\partial y}. $$

$\endgroup$
  • $\begingroup$ So if I figured it out correctly I'll have $$f^{'}_x=\frac{\partial\varphi}{\partial u}y + \frac{\partial\varphi}{\partial v}\frac{1}{y}$$ , where $u=xy$ and $v=\frac{x}{y}$. So for the second derivative I'll have $$\frac{\partial}{\partial x}(\frac{\partial\varphi}{\partial u}y + \frac{\partial\varphi}{\partial v}\frac{1}{y})=\frac{\partial^2 \varphi}{\partial u \partial u}y^2 + \frac{\partial^2 \varphi}{\partial v \partial v}\frac{1}{y^2}+2\frac{\partial^2 \varphi}{\partial u \partial v}$$ Is this right or am I wrong again? $\endgroup$ – randomname Jun 20 '14 at 16:28
  • $\begingroup$ That seems right for the second derivative with respect to $x$. The other second derivatives $\frac{\partial^2 f}{\partial y^2}$ and $\frac{\partial^2 f}{\partial x\partial y}$ are a little more complicated because you have to use the product rule as well as the chain rule. $\endgroup$ – p.s. Jun 20 '14 at 22:02

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.