Use the Chain Rule to find the indicated partial derivatives.

Question

roadjester · Answer

$$\large z=x^2+xy^3, x=uv^2+w^3, y=u+ve^w; $$
I can't find that squiggly d that means partial but I'm looking for dd/du, dz/dv, and dz/dw when u=2, v=1, and w=0.

roadjester · Answer

oops, that dd/du is supposed to be dz/du

anonymous · Answer

WHAT IS PARTIAL DERIVATIVE REFFERED TO AS .@roadjester

cherylim23 · Answer

There's two ways you can do this. (1) Sub x and y into z, and differentiate accordingly or (2) Use the chain rule to differentiate. Looks like you have to use method (2) in this case, but you can always use method (1) to check.

I'll do the first one for you.
dz/du
= dz/dx * dx/du + dz/dy*dy/du
= (2x+y^3)*(v^2) + (3xy^2)*(1)
= [2(u*v^2+w^3)]*(v^2) + 3[(u*v^2+w^3)(u+ve^w)^2]
= 2v^2(u*v^2+w^3) + 3(u*v^2+w^3)(u+ve^w)^2

For the second one, dz/dv, just use dz/dv = dz/dx*dx/dv + dz/dy*dy/dv
For the third one, dz/dw, just use dz/dw = dz/dx*dx/dw + dz/dy*dy/dw

roadjester · Answer

@Ruchii. A partial derivative is when you are taking the derivative of a function that has more than one independent variable. For example, instead of $$y=f(x)$$, you have $$z=f(x,y)$$

In this situation, you can take a partial derivative where you hold one variable constant.

@cherylim23 Thanks, but I think I got the gist. Plus, with a test every Tuesday, I can't help but not catch on to the material. I really hate intersession.

cherylim23 · Answer

No worries, hope that helped