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

• Oct 19th 2009, 03:43 PM
kodos
[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.
• Oct 20th 2009, 12:20 PM
kodos
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.
• Oct 21st 2009, 07:51 AM
kodos
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.