Question

In Exercises 3–10, use the Chain Rule to calculate the partial derivatives. Express the answer in terms of the independent variables. 3. ∂f ∂s , ∂f ∂r ; f (x, y, z) = xy + z2, x = s2, y = 2rs, z = r2 4. ∂f ∂r , ∂f ∂t ; f (x, y, z) = xy + z2, x = r + s − 2t, y = 3rt, z = s2

Answer 1

I'll use subscript notation for brevity, i.e.

$\dfrac{\partial f}{\partial x}=f_x$

3.

$f(x,y,z)=xy+z^2\implies\begin{cases}f_x=y\\f_y=x\\f_z=2z\end{cases}$

$x(r,s)=s^2\implies\begin{cases}x_r=0\\x_s=2s\end{cases}$

$y(r,s)=2rs\implies\begin{cases}y_r=2s\\y_s=2r\end{cases}$

$z(r,s)=r^2\implies\begin{cases}z_r=2r\\z_s=0\end{cases}$

By the chain rule,

$f_r=f_xx_r+f_yy_r+f_zz_r=2xs+4zr=2s^3+4r^3$

$f_s=f_xx_s+f_yy_s+f_zz_s=2ys+2xr=6rs^2$

4.

$x(r,s,t)=r+s-2t\implies\begin{cases}x_r=1\\x_s=1\\x_t=-2\end{cases}$

$y(r,s,t)=3rt\implies\begin{cases}y_r=3t\\y_s=0\\y_t=3r\end{cases}$

$z(r,s,t)=s^2\implies\begin{cases}z_r=0\\z_s=2s\\z_t=0\end{cases}$

By the chain rule,

$f_r=f_xx_r+f_yy_r+f_zz_r=y+3xt=3rt+3r+3s-6t^2$

$f_s=f_xx_s+f_yy_s+f_zz_s=y+4zs=3rt+4s^3$

$f_t=f_xx_t+f_yy_t+f_zz_t=-2y+3xr=3r+3s-12rt$