Function composition by Chain rule

Question

I need to calculate by chain rule the partial derivative of:

dz/du

dz/dv

dz/dw

at point (u,v,w)=(2,1,0) for z=x^2+xy^3, x=uv^2+w^3, y=u+ve^w

I don't know how to find the function composition.

Answer 1

I give you one example and leave the rest for you to practice on.

$z = x^2+xy^3, x=uv^2+w^3, y=u+ve^w\\ Note:\\ \hspace {9 mm} {\partial x \over \partial u} = v^2\\ \hspace{10 mm} {\partial x \over \partial v} = 2uv\\ \hspace{10 mm} {\partial x \over \partial w} = 3w^2\\ \hspace{10 mm} {\partial y \over \partial u} = 1\\ \hspace{10 mm} {\partial y \over \partial v} = e^w\\ \hspace{10 mm} {\partial y \over \partial w} = ve^w\\ $

$$\begin{align} {\partial z \over \partial u} &= {\partial z \over \partial x}\cdot{\partial x \over \partial u} + {\partial z \over \partial y}\cdot{\partial y \over \partial u}\\ &=(2 x + y^3)(v^2) + 3xy^2(1)\hspace{2mm}\text{plugging in the parameters for x and y we get,}\\ &=[2(uv^2+w^3)+(u+ve^w)^3](v^2)+3(uv^2+w^3)(u+ve^w)^2\hspace{3mm}\text{now evaluate at (2,1,0)}\\ &=[2(2\cdot1^2+0^3)+(2+1e^0)^3]1^2+3(2\cdot1^2+0^3)(2+1\cdot e^0)^2\\ &=[2(2)+3^3]+3(2)(3)^2\\ &=31+54=85 \end{align}$$