Question

Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals space 3 u minus v, f subscript x (6 comma 1 )equals 3, and f subscript y (6 comma 1 )equals negative 1. Evaluate fraction numerator partial differential z over denominator partial differential v end fraction at (u comma v )equals (1 comma 2 ). **Note 1: Your answer will be an integer.

Answer 1

`z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}`

We're given that
`f_x(6,1)=3` and
`f_y(6,1)=-1`, and want to find
`(\partial z)/(\partial v)(1,2)`.

By the chain rule, we have

`(\partial z)/(\partial v)=(\partial z)/(\partial x)(\partial x)/(\partial v)+(\partial z)/(\partial y)(\partial y)/(\partial v)`

and

`(\partial x)/(\partial v)=2v`

`(\partial y)/(\partial v)=-1`

Then

`(\partial z)/(\partial v)(1,2)=(\partial z)/(\partial x)(6,1)(\partial x)/(\partial v)(1,2)+(\partial z)/(\partial y)(6,1)(\partial y)/(\partial v)(1,2)`

(because the point
`(x,y)=(6,1)` corresponds to
`(u,v)=(1,2)`)

`\implies(\partial z)/(\partial v)(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}`