partial derivatives and the chain rule

[peacefreak77]

Use the chain rule to find the partial derivative of z with respect to s and the partial derivative of z with respect to t where
z = x^2 + xy + y^2, x = 3 s + 4 t, y = 5 s + 6 t

Okay. Although it is inaccurate, for convenience's sake let's say every time I type dx/dy (etc) I mean the partial derivative of x with respect to y.

I figured there were two pathways to get from z to s and two pathways to get from z to t (z-->x-->s, z-->y-->s, z-->x-->t, z-->y-->t). Therefore dz/ds=dz/dx*dx/ds+dz/dy*dy/ds and dz/dt=dz/dx*dx/dt+dz/dy*dy/dt.

I computed dz/dx to be 2x+y, dz/dy to be x+2y, dx/ds to be 3, dx/dt to be 4, dy/ds to be 5, and dy/dt to be 6.

Plugging these into my equations I found dz/ds to be 11x+13y and dz/dt to be 14x+16y.

My problem is that the partial derivatives with respect to s need to be in terms of s, so somehow I need to get 11x+13y and 14x+16y in terms of s. I tried to do this by using linear combination on the two equations provided that had the variables of s, x, and y, eliminating t, but it didn't work.

I'm kind of stuck at the end of all this. Maybe it's just an algebra problem, maybe I have to throw in another partial derivative, but I don't know what to do.

 

[galactus]

Let's see. These can be kinda confusing, but it's just the chain rule.

\(\displaystyle \L\\\frac{\partial{z}}{\partial{s}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{s}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}\)

\(\displaystyle \L\\\frac{\partial{z}}{\partial{x}}=2x+y\)

\(\displaystyle \L\\\frac{\partial{x}}{\partial{s}}=3\)

\(\displaystyle \L\\\frac{\partial{z}}{\partial{y}}=2y+x\)

\(\displaystyle \L\\\frac{\partial{y}}{\partial{s}}=5\)


\(\displaystyle \L\\\frac{\partial{z}}{\partial{s}}=(2x+y)(3)+(2y+x)(5)=11x+13y\)

\(\displaystyle \L\\=11(3s+4t)+13(5s+6t)=98s+122t\)

Check me out. It's easy to walk down the primrose path on these problems.

 

[peacefreak77]

that is the correct answer, but how did you get the x and y in terms of s and t?

Sorry if I'm missing something really obvious here...

 

[galactus]

Yes, you are missing the obvious. :lol:

You were given x=3s+4t and y=5s+6t. Just sub them in.

 

[peacefreak77]

thank you!!! wow that was really stupid of me...

I can do the chain rule with partial derivatives, I just can't substitute in given equations...