# Trouble with partial derivatives

I've no clue how to get started .I'am unable to even understand what the hint is saying.I need your help please.

Given $$u = f(ax^2 + 2hxy + by^2), \qquad v = \phi (ax^2 + 2hxy + by^2),$$ then prove that
$$\frac{\partial }{\partial y} \left ( u\frac{\partial u }{\partial x} \right ) = \frac{\partial }{\partial x}\left ( u \frac{\partial v}{\partial y} \right ).$$

Hint. Given $$u = f(z),v = \phi(z), \text{where} z = ax^2 + 2hxy + by^2$$

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You have already posted 11 questions, 4 of which with at least 4 answers. You have accepted none. –  Américo Tavares Sep 5 '11 at 13:44
To accept an answer: Choose the answer that helped you the most. You will find a tick mark (a V shape) below the vote count for the answer. Click it. And, I am not sure I am allowed to point it out. But you have cast zero votes till now. Usually I would say how you vote is your business, but this behavior is not healthy, and is even considered rude by users (including me) answering your question. –  Srivatsan Sep 5 '11 at 13:55
alok, Responders get reputation because people vote to show that his/her solution is good. And, votes have nothing to do with who does your question first; this is not a talent competition for us. Finally, votes are not just for me to gain more reputation. They are how people can differentiate good solutions from bad; in short, votes are how this site works. From that point of view, I argue that proper voting is one's duty in this site. I just thought this had to be pointed out by somebody. I will say no more on this. (I don't mean this personally as well, sorry if you take it so.) –  Srivatsan Sep 5 '11 at 14:06
@alok: Why should you delete all other posts (anyway, you can't as soon as an answer has been posted)? Just learn how to accept the answer. Open one of your questions. Below the down arrow under the votes there is a checkmark (this is true for any answer). Just click the checkmark near the answer you want. It's very simple. This site is not only for you. Why are you so frustrated because you need to do something in exchange for receiving answers to your questions? –  Beni Bogosel Sep 5 '11 at 14:19
Open one of your questions. Scroll down to an answer. Look at the column on the left. Starting from the top, you first meet an up-arrow (click on this to upvote the answer) then a score (the number of upvotes minus the number of downvotes the answer received) then a down-arrow (click on this to downvote the answer) then a checkmark (click on this to accept the answer). Do you see the checkmark? –  Did Sep 5 '11 at 14:57

I recommend you to use the chain rule, i.e. given functions $f:U\mathbb{R}^n\rightarrow V\subset\mathbb{R}^m$ differenciable on $x$ and $g:V_0\subset V\subset\mathbb{R}^m\rightarrow W\subset\mathbb{R}^p$ differenciable on $f(x)$ we have $$D_x(g\circ f)=D_{f(x)}(g)D_x(f)$$ where $D_x(f)$ represents the Jacobian matriz of $f$ at $x$.

In your particular case when $n=2$ and $m=p=1$, we have for each coordinate that $$\left.\frac{\partial g\circ f}{\partial x_i}\right|_{(x_1,x_2)}=\left.\frac{d\,g}{dx}\right|_{f(x_1,x_2)}\left.\frac{\partial f}{\partial x_i}\right|_{(x_1,x_2)}$$

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I think that the hint and chain rule are more than enough. There may be a typo in your question since in the RHS you have both $u,v$ in the derivation and in the LHS you have only $u$.
$$u=f(z) \Rightarrow \frac{\partial u}{\partial x}=f'(z)\cdot \frac{\partial z}{\partial x}=f'(z)\cdot(2ax +2hy)$$
Now to compute everything in the LHS you just plug in $v$ and the previous result and use the product rule.
$$\frac{\partial}{\partial y}\left(f(z) \cdot f'(z) \cdot (2ax+2hy)\right),$$