Let $p=y^2$, $q=xy$, $r=-x^2$ and $u=f(p,q,r)$ where $f$ has continuous second order partial derivatives. Calculate $u_{xy}$ in terms of $p,q,r$ and partial derivatives of $f$ with respect to these variables.

Anyone can guide me for this kind of question so that I can do for all similar kinds?

My final result is : $$\frac{2y^2d^2(u)}{dqdp} -\frac{4xyd^2 (u)}{drdp} + \frac{du}{dq} +\frac{xyd^2(u)}{dq^2} -\frac{2x^2d^2(u)}{drdq}$$

  • $\begingroup$ Do you know implicit differentiation? $\endgroup$ – Archis Welankar Nov 20 '15 at 5:18
  • $\begingroup$ Your final result has $s,t$ in it which were not there before. What are they? $\endgroup$ – Ross Millikan Nov 20 '15 at 5:46
  • $\begingroup$ Hint: you have $u=f(p,q,r)=f($ some expressions in $x$ and $y$) from substituting for the variables. You are expected to use the chain rule to get an answer in things like $f_{pq}$ Where is your problem? $\endgroup$ – Ross Millikan Nov 20 '15 at 5:58
  • $\begingroup$ @RossMillikan sorry for the typo $\endgroup$ – UnusualSkill Nov 20 '15 at 7:01

You have $u=f(y^2,xy,-x^2)$. Then $u_x=y\frac {\partial f(p,q,r)}{\partial q}-2x\frac {\partial f(p,q,r)}{\partial r}$ by the chain rule. Now take the partial with respect to $y$


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