Answer
(a) $\frac{{\partial f}}{{\partial x}} = \cos \left( y \right)$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = - x\sin \left( y \right)$
(b) $\frac{{\partial f}}{{\partial v}} = 2v\cos \left( y \right) + x\sin \left( y \right)$
(c) For $\left( {u,v} \right) = \left( {2,1} \right)$, we get $\left( {x,y} \right) = \left( {5,1} \right)$
$\frac{{\partial f}}{{\partial v}}{|_{\left( {u,v} \right) = \left( {2,1} \right)}} = 2\cos \left( 1 \right) + 5\sin \left( 1 \right)$
Work Step by Step
(a) We have $f\left( {x,y} \right) = x\cos \left( y \right)$ and $x = {u^2} + {v^2}$ and $y = u - v$.
The primary derivatives are
$\frac{{\partial f}}{{\partial x}} = \cos \left( y \right)$, ${\ \ }$ $\frac{{\partial f}}{{\partial y}} = - x\sin \left( y \right)$
Calculate
$\frac{{\partial x}}{{\partial v}} = 2v$, ${\ \ }$ $\frac{{\partial y}}{{\partial v}} = - 1$
(b) Using the Chain Rule we calculate:
$\frac{{\partial f}}{{\partial v}} = \frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial v}} + \frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial v}}$
$\frac{{\partial f}}{{\partial v}} = 2v\cos \left( y \right) + x\sin \left( y \right)$
(c) Since $x = {u^2} + {v^2}$ and $y = u - v$. For $\left( {u,v} \right) = \left( {2,1} \right)$, we get
$\left( {x,y} \right) = \left( {5,1} \right)$
From part (b) we obtain $\frac{{\partial f}}{{\partial v}} = 2v\cos \left( y \right) + x\sin \left( y \right)$. So,
$\frac{{\partial f}}{{\partial v}}{|_{\left( {u,v} \right) = \left( {2,1} \right)}} = 2\cos \left( 1 \right) + 5\sin \left( 1 \right)$
