Question

1. use the multi - variable chain rule to calculate the derivatives. express your answer in terms of the independent variables. next, compose the variables first before differentiating, and then calculate the derivatives. (observe that the second method is usually less annoying...) (a) $f(x,y)=x^{2}+y^{3}$, $x = rs^{2}$, $y = st^{2}$. $\frac{partial f}{partial r},\frac{partial f}{partial s},\frac{partial f}{partial t}$ (b) $f(x,y,z)=xsin(y + z^{2})$, $x = st$, $y = s^{2}$, $z=cos t$. $\frac{partial f}{partial s},\frac{partial f}{partial t}$

Explanation:

Step1: Recall the multi - variable chain rule

The multi - variable chain rule for $\frac{\partial f}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$ and $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$ for a function $f(x,y)$ where $x = x(s,t)$ and $y = y(s,t)$. For $f(x,y,z)$ with $x = x(s,t),y = y(s,t),z = z(s,t)$, $\frac{\partial f}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial s}$ and $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial t}$.

Step2: Calculate partial derivatives for part (a)

Given $f(x,y)=x^{2}+y^{3}$, $x = rs^{2}$, $y = st^{2}$.
First, find $\frac{\partial f}{\partial x}=2x$, $\frac{\partial f}{\partial y}=3y^{2}$, $\frac{\partial x}{\partial r}=s^{2}$, $\frac{\partial x}{\partial s}=2rs$, $\frac{\partial y}{\partial r}=0$, $\frac{\partial y}{\partial s}=t^{2}$.
$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=2x\cdot s^{2}+3y^{2}\cdot0 = 2x s^{2}=2(rs^{2})s^{2}=2r s^{4}$.
$\frac{\partial f}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=2x\cdot2rs + 3y^{2}\cdot t^{2}=4xrs+3y^{2}t^{2}=4(rs^{2})rs + 3(st^{2})^{2}t^{2}=4r^{2}s^{3}+3s^{2}t^{6}$.
$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}=2x\cdot0+3y^{2}\cdot2st=6y^{2}st = 6(st^{2})^{2}st=6s^{3}t^{5}$.

Step3: Calculate partial derivatives for part (b)

Given $f(x,y,z)=x\sin(y + z^{2})$, $x = st$, $y = s^{2}$, $z=\cos t$.
$\frac{\partial f}{\partial x}=\sin(y + z^{2})$, $\frac{\partial f}{\partial y}=x\cos(y + z^{2})$, $\frac{\partial f}{\partial z}=2xz\cos(y + z^{2})$.
$\frac{\partial x}{\partial s}=t$, $\frac{\partial x}{\partial t}=s$, $\frac{\partial y}{\partial s}=2s$, $\frac{\partial y}{\partial t}=0$, $\frac{\partial z}{\partial s}=0$, $\frac{\partial z}{\partial t}=-\sin t$.
$\frac{\partial f}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial s}=\sin(y + z^{2})\cdot t+x\cos(y + z^{2})\cdot2s+2xz\cos(y + z^{2})\cdot0=t\sin(s^{2}+\cos^{2}t)+2st\cos(s^{2}+\cos^{2}t)$.
$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial t}=\sin(y + z^{2})\cdot s+x\cos(y + z^{2})\cdot0+2xz\cos(y + z^{2})\cdot(-\sin t)=s\sin(s^{2}+\cos^{2}t)-2st\cos t\sin t\cos(s^{2}+\cos^{2}t)$.

Answer:

(a) $\frac{\partial f}{\partial r}=2rs^{4}$, $\frac{\partial f}{\partial s}=4r^{2}s^{3}+3s^{2}t^{6}$, $\frac{\partial f}{\partial t}=6s^{3}t^{5}$
(b) $\frac{\partial f}{\partial s}=t\sin(s^{2}+\cos^{2}t)+2st\cos(s^{2}+\cos^{2}t)$, $\frac{\partial f}{\partial t}=s\sin(s^{2}+\cos^{2}t)-2st\cos t\sin t\cos(s^{2}+\cos^{2}t)$