## Calculus (3rd Edition)

Given $$R(x, y)=(3 x+4 y)^{5},\ \ \ x=u^{2},\ \ \ y=u v$$ Since \begin{align*} \frac{\partial R}{\partial x}&=15(3x+4y)^4 \\ \frac{\partial R}{\partial y}&= 20(3x+4y)^4 \end{align*} and \begin{align*} \frac{\partial x}{\partial u}&=2u \ \ \ \ \ \ \frac{\partial x}{\partial v}=0 \\ \frac{\partial y}{\partial u}&=v \ \ \ \ \ \ \ \ \ \frac{\partial y}{\partial v}= u \end{align*} Then \begin{align*} \frac{\partial R}{\partial u}&=\frac{\partial R}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial R}{\partial y}\frac{\partial y}{\partial u} \\ &=30u(3x+4y)^4 +20v(3x+4y)^4 \\ &=30u(3u^2+4uv)^4 +20v(3u^2+4uv)^4 \\ \frac{\partial R}{\partial v}&=\frac{\partial R}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial R}{\partial y}\frac{\partial y}{\partial v} \\ &= 20u(3x+4y)^4\\ &= 20u(3u^2+4uv)^4 \end{align*}