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 $\frac{\partial z}{\partial v}(1,2)$.

By the chain rule, we have

$\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}$

and

$\dfrac{\partial x}{\partial v}=2v$

$\dfrac{\partial y}{\partial v}=-1$

Then

$\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)$

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

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