\section{Partial Derivatives}
\begin{enumerate}[1.]


\item{\it 
Let
$$f(x,y)=\left\{\begin{matrix}
\displaystyle{\frac{\sin^3 x}{x^2 + y^2}}&\hbox{ if }(x,y)\neq (0,0)\\[2ex]
0&\hbox{ if }(x,y)=(0,0)\\
\end{matrix}
\right.
$$
Find the first order partial derivatives of $f$ at $(0,0)$ using the definition of partial derivatives.
}

\bigskip
Using that $\lim_{x\to 0}\frac{\sin x}{x}=1$, we get
\begin{align*}
f_1(0,0)&=\lim_{x\to 0} \frac{f(x,0)-f(0,0)}{x}
=\lim_{x\to 0} \frac{\sin^3 x-0}{x^3}
\\
&
= \lim_{x\to 0} \Big(\frac{\sin x}{x}\Big)^3 = 1,
\\
& \\
f_2(0,0)&=\lim_{y\to 0} \frac{f(0,y)-f(0,0)}{y}
=\lim_{y\to 0} \frac{0}{y}=0.
\end{align*}





\item
{\it Let 
$$f(x,y)=\left\{\begin{matrix}
\displaystyle{\frac{x^3+xy-y^3}{x^2+y^2}}&\hbox{ if }(x,y)\neq (0,0)\\[2ex]
0&\hbox{ if }(x,y)=(0,0)\\
\end{matrix}
\right.
$$

\bigskip
Find $\displaystyle{\frac{\partial f}{\partial x}(0,0)}$ and 
$\displaystyle{\frac{\partial f}{\partial y}(0,0)}$.}

\bigskip
We have 
$\displaystyle{f(x,0)=\frac{x^3}{x^2}=x}$ and
$\displaystyle{f(0,y)=\frac{-y^3}{y^2}=-y}$ for $x,y\neq 0$, and it is also true that $f(x,0)=0=x$ at $x=0$ and $f(0,y)=0=-y$ at $y=0$. Hence
\begin{align*}
\frac{\partial f}{\partial x}(0,0)&= \frac{\partial}{\partial x}(x)\Big|_{x=0}
=1,\\
\frac{\partial f}{\partial y}(0,0)&=\frac{\partial}{\partial y}(-y)\Big|_{y=0}=-1.
\end{align*}

\bigskip

\end{enumerate}
