# Thread: [SOLVED] chain rule with functions related by equations?

1. ## [SOLVED] chain rule with functions related by equations?

I just don't get this problem, I feel I'm missing something obvious.

Find $\displaystyle dg/du$ if

$\displaystyle 1) x+y= uv$
$\displaystyle 2) xy=u-v$

and

$\displaystyle x=g(u,v)$
$\displaystyle y=h(u,v)$

Poor attempt of solution:

I try to obtain an expression for x as a function of u and v, but I can't seem to be able to get anything from that system of equations.

From 1) I get $\displaystyle x= uv-y$ , but from 2) $\displaystyle y=(u-v)/x$

If I replace into 1) $\displaystyle x=uv-uv/x$

I don't think I'm getting something from that

If $\displaystyle x=g(u,v)$ then

$\displaystyle g(u,v)=uv-y$
$\displaystyle g(u,v)=uv-h(u,v)$

Another thought from there: $\displaystyle dg/du=v-dh/du$

$\displaystyle h(u,v)=y=(u-v)/x$

$\displaystyle dh/du=x$

then,

$\displaystyle dg/du=v-x$ , but $\displaystyle x=g(u,v)$

and I think I'm running in circles here.

2. What if I tried to obtain a F(x,y)= 0 which I could then derive and get an expression for dg/du? The problem would be getting rid of u and v, but the truth is I suck at solving systems of equations. Maybe adding and multiplying the equations...

Another idea would be that y could be defined implicitly as a function of x.
It would be like F(x,y)=F(x,f(x))=0, where x is defined by u and v and keeps holding the condition.

3. I finally got the resolution for this exercise. The idea was derive both equations in terms of u. Then, having a system of equations, solve for dg/du.