0

Suppose I have the functions $f(x,y)=xy-y$ and $g(u,v)=f(u,uv)$ and I want to calculate dg(a,b) two ways (1) explicitly and (2) using chain rule

My attempt: for (1) it seems easy enough just plug x=u and y=uv then, $g(u,v)=f(u,uv)=u^2v-uv$ and $\nabla g =(2uv-v,u^2-u)$ and plug points (a,b) for (u,v)

for (2) I know that, $\frac{\partial g}{\partial u}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial u}=\frac{\partial g}{\partial x}(1)+\frac{\partial g}{\partial y}(v)$ but how do I proceed from here? I should get the same answer as part (1) no?

pendulumbob
  • 3

1 Answers1

0

$$\begin{align} \text{let } x & = u \qquad \therefore \frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 0 \\[2ex] \text{let } y &= uv \qquad \therefore \frac{\partial y}{\partial u} = v , \quad \frac{\partial y}{\partial v} = u \\[2ex] g(u, v) & = g(x, y/x) = f(x,y) = xy-y \\[2ex] \frac{\partial g}{\partial x} & = y \qquad\qquad= uv \\[2ex] \frac{\partial g}{\partial y} & = x-1 \qquad = u-1 \end{align}$$

Put them together to obtain $$\frac{\partial g}{\partial u} = \frac{\partial g}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial g}{\partial y}\frac{\partial y}{\partial u} \\[2ex] \dfrac{\partial g}{\partial v} = = \frac{\partial g}{\partial x}\frac{\partial x}{\partial v}+ \frac{\partial g}{\partial y}\frac{\partial y}{\partial v}$$

Graham Kemp
  • 129,094