3
$\begingroup$

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

  • 2
    *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?2011-09-05

2 Answers 2

2

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 $\left.\frac{\partial g\circ f}{\partial x_i}\right|_{(x_1,x_2)}=\left.\frac{d\,g}{dx}\right|_{f(x_1,x_2)}\left.\frac{\partial f}{\partial x_i}\right|_{(x_1,x_2)}$

3

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),

and again, use chain rule whenever necessary.