Thread: Multivariable Chain Rule Help!

1. Multivariable Chain Rule Help!

Suppose f is a differentiable function of x and y, and g(r,s) = f(2r-s,s^2-4r). Calculate dg/dr at (1,2) and dg/ds at (1,2) given that f(1,2)=6, g(1,2)=3, df/dx(1,2)=2 and df/dy(1,2)=5.

Thanks for reading my question!

2. bump, someone help please?

3. Originally Posted by MathTooHard
Suppose f is a differentiable function of x and y, and g(r,s) = f(2r-s,s^2-4r). Calculate dg/dr at (1,2) and dg/ds at (1,2) given that f(1,2)=6, g(1,2)=3, df/dx(1,2)=2 and df/dy(1,2)=5.

Thanks for reading my question!

Put x = 2r - s, y = s^2 - 4r ==> g(r,s) = f(x(r,s),y(r,s)) ==>

dg/dr = (df/dx)(dx/dr) + (df/dy)(dy/dr) = (df/dx)*2 + (df/dy)*(-4), and since at (1,2) we have (df/dx)(1,2) = 2 and (df/dy)(1,2) = 5, we get....well, now you end the argument.

Tonio

4. Thanks.

What about dg/ds? That s^2 term is giving me bid headaches.

5. Originally Posted by MathTooHard
Thanks.

What about dg/ds? That s^2 term is giving me bid headaches.

It's EXACTLY similar as dg/dr but with s!

Tonio

6. I get -2+10s for dg/ds. From here, do I plug in 2 for s?