\section{Implicit Functions}
\begin{enumerate}[1.]
\item 
{\it A surface is given by the equation $e^x(y+1)-e^y z = (e^z-1)x$.

\medskip

(a) Prove that the point $P=(1,0,1)$ lies on the surface.}

\medskip

We check that $(x,y,z)=(1,0,1)$ satisfies the equation of the surface: $e^1(0+1)-e^0\cdot 1 = (e^1-1)\cdot 1$, $e-1=e-1$, which is true.

\medskip



{\it (b) 
Find $\displaystyle{\frac{\partial z}{\partial x} }$ and $\displaystyle{ \frac{\partial z}{\partial y} }$ at $P$. }

\medskip

We rewrite the equation of the surface as $F(x,y,z)=0$, where 
$F(x,y,z)=e^x(y+1)-e^y z - (e^z-1)x$. Then $F$ has continuous partial derivatives of all orders, and
\begin{align*}
& F_x(x,y,z)=e^x(y+1)-(e^z-1)\\
& F_y(x,y,z)=e^x-e^yz\\
& F_z(x,y,z)=-e^y-e^z x\\
\end{align*}
so that
\begin{align*}
&F_x(1,0,1)= e^1(0+1)-(e^1- 1) =1\\
&F_y(1,0,1)= e^1-e^0\cdot 1 =e-1\\
&F_z(1,0,1)= -e^0 - e^1\cdot 1 =-1-e
\end{align*}
 Since $F_z(1,0,1)\neq 0$, the equation defines $z$ as a function of $x,y$ in a neighbourhood of $P$, and we have at $P$
$$
\frac{\partial z}{\partial x} =-\frac{F_x(1,0,1)}{F_z(1,0,1)}=\frac{1}{e+1},
\ \ 
\frac{\partial z}{\partial y} =-\frac{F_y(1,0,1)}{F_z(1,0,1)}=\frac{e-1}{e+1}.
$$

\bigskip

\item {\it Suppose that $z$ is a function of $x,y$ such that $\sin(xz)-yz=0$. Find
$\frac{\partial{z}}{\partial x}$ and $\frac{\partial{z}}{\partial y}$ where possible.}


\medskip

We first differentiate the equation in $x$:
$$
\cos(xz)\left( z+ x\frac{\partial z}{\partial x} \right) - y  \frac{\partial z}{\partial x}=0
$$
$$
\frac{\partial z}{\partial x} \left( x\cos(xz)-y \right) = -z \cos(xz)
$$
$$
\frac{\partial z}{\partial x} = - \frac{ z \cos(xz)}{ x\cos(xz)-y}
$$
Now differentiate in $y$:
$$
\cos(xz)\left( x\frac{\partial z}{\partial y} \right) - z -y  \frac{\partial z}{\partial y}=0
$$
$$
\frac{\partial z}{\partial y} \left( x\cos(xz)-y \right) = z
$$
$$
\frac{\partial z}{\partial y} =  \frac{ z}{ x\cos(xz)-y}
$$
Both derivatives exist whenever $x\cos(xz)-y\neq 0$.

\bigskip

\item
{\it Suppose that the functions $x(u,v)$ and $y(u,v)$ solve the equations
$$\left\{\begin{matrix}
u & = & 3x+2e^y \\
v & = & e^x - y
\end{matrix}
\right.
$$
in some neighbourhood of the point where $(x,y)=(1,0)$.
Find $\displaystyle{ \frac{\partial x}{\partial u}}$ and $\displaystyle{ \frac{\partial y}{\partial u}}$ where possible, and evaluate them at $(x,y)=(1,0)$.}





\medskip

We differentiate the equations in $u$:
$$\left\{\begin{matrix}
1 & =  &3  \frac{\partial x}{\partial u} +2e^y \frac{\partial y}{\partial u} \\
\\
0 & =  &e^x \frac{\partial x}{\partial u} - \frac{\partial y}{\partial u} \\
\end{matrix}
\right.
$$
From the second equation, $\frac{\partial y}{\partial u} = e^x \frac{\partial x}{\partial u}$.
Plugging this into the first equation, we get
$$
3 \frac{\partial x}{\partial u} +2e^y e^x \frac{\partial x}{\partial u} 
= (3 +e^{x+y} ) \frac{\partial x}{\partial u} =1,
$$
$$
\frac{\partial x}{\partial u} = \frac{1}{3+2e^{x+y}}, \ \ \frac{\partial y}{\partial u}= \frac{e^x}{3+2e^{x+y}}.
$$
Finally, plugging in $(x,y)=(1,0)$, we get that at that point,
$$
\frac{\partial x}{\partial u} = \frac{1}{3+2e}, \ \ \frac{\partial y}{\partial u}= \frac{e}{3+2e}.
$$



\bigskip



\item {\it 
The equations
\begin{align*}
u&=x^3-y^2\\
v&=2xy^3
\end{align*}
define $x,y$ implicitly as functions of $u,v$ near $x=-1$, $y=1$.
Find $\displaystyle \frac{\partial x}{\partial u}$ and $\displaystyle \frac{\partial y}{\partial u}$
where possible, and evaluate them at 
$x=-1$, $y=1$.}

\medskip


Differentiating the two equations with respect to $u$, we get
\begin{align*}
1&= 3x^2 \frac{\partial x}{\partial u} - 2y \frac{\partial y}{\partial u}\\
0&= 2y^3 \frac{\partial x}{\partial u} + 6xy^2 \frac{\partial y}{\partial u}
\end{align*}
so that
$$
\frac{\partial x}{\partial u} 
=\frac{
\left|\begin{matrix}
1 & -2y \\
0 & 6xy^2 \\
            \end{matrix}\right|
}
{
\left|\begin{matrix}
3x^2  & -2y \\
2y^3  & 6xy^2 \\
            \end{matrix}\right|
}
=\frac{6xy^2}{18x^3y^2+4y^4}=\frac{3x}{9x^3+2y^2}
$$
$$
\frac{\partial y}{\partial u} 
=\frac{
\left|\begin{matrix}
3x^2 & 1 \\
2y^3 & 0  \\
            \end{matrix}\right|
}
{
\left|\begin{matrix}
3x^2  & -2y \\
2y^3  & 6xy^2 \\
            \end{matrix}\right|
}
=\frac{-2y^3}{18x^3y^2+4y^4}=\frac{-y}{9x^3+2y^2}
$$
At $(x,y)=(-1,1)$, we have 
$$
\frac{\partial x}{\partial u}=\frac{-3}{-9+2}=\frac{3}{7},\ \ 
\frac{\partial y}{\partial u}=\frac{-1}{-9+2}=\frac{1}{7}.$$


\bigskip


\item
{\it The quantities $u,v$ are defined as functions of $x,y$ so that they satisfy the equations
$x=u\cos(u+v)$, $y=v\sin u$.
Find $\displaystyle \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x}$ at the point where $ u=\pi/3$, $v=0$. (Hint: $\sin(\pi/3)=\frac{\sqrt{3}}{2}$,
$\cos(\pi/3)=\frac{1}{2}$.)}

\medskip


Differentiating the equations $x=u\cos(u+v)$, $y=v\sin u$ in $x$, we get
\begin{align*}
&1=u_x \cos(u+v) - u \sin(u+v)\cdot (u_x+v_x),
\\
&0=v_x \sin u + v \cos u \cdot u_x.
\end{align*}
At $u=\pi/3$, $v=0$, we get
\begin{align*}
&1=u_x \cos(\pi/3) - \frac{\pi}{3} \sin(\pi/3)\cdot (u_x+v_x) = 
\frac{u_x}{2}  - \frac{\pi\sqrt{3} }{6} (u_x+v_x),\\
&0=v_x \sin (\pi/3) =\frac{\sqrt{3}}{2}v_x.
\end{align*}
Hence $v_x=0$, and 
$$
u_x=\frac{1}{\frac{1}{2}  - \frac{\pi\sqrt{3} }{6} } = \frac{6}{3-\pi\sqrt{3}}.
$$
\end{enumerate}
