\section{Implicit Functions}
The Implicit Function Theorem is often stated in textbooks in a general form that uses vector-valued functions. In this section, we focus on two important special cases that cover most of the applications we need, and mention vector-valued functions briefly at the end. We assume that all functions involved are $C^1$ on their domain of definition, and ask the following two questions.
\begin{enumerate}
\item Given an equation $F(x,y) = 0$, does this define $y$ as a function of $x$? Similarly, given that $F(x,y,z) = 0$, does this define $z$ as a function of $x$ and $y$? (Of course, we could keep asking this for functions of $n$ variables).

\medskip

\item Given a system of equations
$$
\begin{cases}
u = f(x,y) \\
v = g(x,y)
\end{cases}
$$
can we solve for $x$ and $y$ as functions of $u$ and $v$? A common example of this is change of variables---we will use this for example in multivariable integration.
\end{enumerate}
In both situations, we are especially interested in whether the resulting functions are differentiable---and, if so, is there a formula for calculating their derivative?

\subsection{Single Equation with Implicitly Defined Function} Consider the equation $F(x,y) = 0$ in two variables.  Recall that we are attempting to determine whether the variable $y$ can be written as a function of $x$ based on this equation. So, suppose we had $F(x,y(x)) = 0$. We can differentiate the equation using the Chain Rule to obtain
$$
0 = \frac{d}{dx} F(x, y(x)) = F_1 \big(x, y(x)\big) + F_2 \big( x, y(x) \big) \frac{dy}{dx}.
$$
If we make the assumptions that $F$ is a $C^1$ function and $F_2 \neq 0$, we may solve the previous equation for $\frac{dy}{dx}$.  We obtain
\begin{equation}\label{eq:34eq1}
\frac{dy}{dx} = - \frac{F_1}{F_2}
\end{equation}
which gives the variable $y$ as an \underline{implicit function} of the variable $x$. You may have seen this version of implicit differentiation in single-variable calculus.

\begin{example}
Given that $F(x,y) = x^2 + y^2 - 4 = 0$ (which is an equation relating the variables $x$ and $y$), let us determine $\frac{dy}{dx}$.  From equation \eqref{eq:34eq1}, we know that
$$
\frac{dy}{dx} = - \frac{F_1 (x,y)}{F_2 (x,y)} = - \frac{2x}{2y} = - \frac{x}{y},
$$
which is well-defined so long as $F_2 (x,y) = y \neq 0$. 
\end{example}



\begin{figure}[H]
\centering
\includegraphics[width= .5\textwidth]{34fig1.png}
\caption{The graph of the equation $F(x,y) = 0$ in the above example. }
\end{figure}


Notice that $\frac{dy}{dx}$ can be a function of both $x$ and $y$. In the above example, the slopes of the curve (which is a circle) at the points $P$ and $Q$ are different.even though their $x$-coordinates are the same.  Moreover,  at the points where $y = 0$, $y$ is not well defined as a function of $x$. So, our method for determining $\frac{dy}{dx}$ fails here. 

\medskip
For the equation $F(x,y,z) = 0$, the method is largely similar to what we just saw.  Suppose $z = z(x,y)$ is some function of $x$ and $y$ which satisfies the equation. Then we can find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ by differentiating implicitly the equation $F(x,y,z(x,y))=0 $ and using the Chain Rule.  Since $F(x,y,z(x,y))$ depends upon $x$ in both the $x$ and $z(x,y)$ variables, the Chain Rule gives
$$
0 = \frac{\partial}{\partial x} \big[F(x,y, z(x,y)) \big] = F_1 (x,y,z) + F_3 (x,y,z(x,y)) \cdot \frac{\partial z}{\partial x},
$$
which, after solving for $\frac{\partial z}{\partial x}$, gives
\begin{equation}\label{eq:34eq2}
\frac{\partial z}{\partial x} = - \frac{F_1 (x,y,z)}{F_3 (x,y,z)},  \quad \forall (x,y,z) \textrm{ with } F_3 (x,y,z) \neq 0.
\end{equation}
where, again, we have assumed that $F$ is $C^1$.  Similar reasoning gives,
\begin{equation}\label{eq:34eq3}
\frac{\partial z}{\partial y} = - \frac{F_2 (x,y,z)}{F_3 (x,y,z)}, \quad \forall (x,y,z) \textrm{ with } F_3 (x,y,z) \neq 0.
\end{equation}
We will see some examples illustrating the usage of \eqref{eq:34eq2} and \eqref{eq:34eq3} in the Worked Problems Section. 

\subsection{Systems of Equations}
Now suppose that we are given the system of equations
$$
\begin{cases}
u = f(x,y) \\
v = g(x,y)
\end{cases}
$$
Can we solve for $x$ and $y$ as \underline{differentiable} functions of $u$ and $v$? If so, what are the formulae for the derivatives of $x$ and $y$ in terms of $u$ and $v$?

Writing $x = x(u,v)$ and $y = y (u,v)$, and assuming that $x$ and $y$ are differentiable,  we differentiate in $u$ and use the Chain Rule to write
\begin{eqnarray*}
1 & = & \frac{\partial }{\partial u} u = \frac{\partial}{\partial u} \big[f(x(u,v), y(u,v)) \big] = f_1 (x,y) \frac{\partial x}{\partial u} + f_2 (x,y) \frac{\partial y}{\partial u} \\[2ex]
0 & = & \frac{\partial}{\partial u} v = \frac{\partial }{\partial u} \big[g(x(u,v), y(u,v)) \big] = g_1 (x,y) \frac{\partial x}{\partial u} + g_2 (x,y) \frac{\partial y}{\partial u}
\end{eqnarray*}
Re-writing this equality in matrix form, we have
\begin{equation}\label{eq:34eq4}
\begin{pmatrix}
1 \\
0
\end{pmatrix}
=
\begin{pmatrix}
f_1 & f_2 \\
g_1 & g_2 
\end{pmatrix}
\begin{pmatrix}
x_u \\
y_u
\end{pmatrix}
\end{equation}
where $x_u = \frac{\partial x}{\partial u}$ and $y_u = \frac{\partial y}{\partial u}$. We solve the system using Cramer's rule, which gives
\begin{eqnarray*}
\frac{\partial x}{\partial u} = \frac{\begin{vmatrix} 1 & f_2 \\ 0 & g_2 \end{vmatrix}}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}} = \frac{g_2}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}} \quad \quad \quad \quad \frac{\partial y}{\partial u} = \frac{\begin{vmatrix} f_1 & 1 \\ g_1 & 0 \end{vmatrix}}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}} = - \frac{g_1}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}}
\end{eqnarray*}



Similarly, if we differentiate in $v$, we arrive at the system of equations
\begin{equation}\label{eq:34eq5}
\begin{pmatrix}
0 \\
1
\end{pmatrix}
=
\begin{pmatrix}
f_1 & f_2 \\
g_1 & g_2
\end{pmatrix}
\begin{pmatrix}
x_v \\
y_v
\end{pmatrix}
\end{equation}
We use Cramer's Rule to solve this system of equations:
\begin{eqnarray*}
\frac{\partial x}{\partial u} = \frac{\begin{vmatrix} 0 & f_2 \\ 1 & g_2 \end{vmatrix}}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}} = \frac{-f_2}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}} \quad \quad \quad \quad \frac{\partial y}{\partial u} = \frac{\begin{vmatrix} f_1 & 0 \\ g_1 & 1 \end{vmatrix}}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}} =  \frac{f_1}{\begin{vmatrix} f_1 & f_2 \\ g_1 & g_2 \end{vmatrix}}
\end{eqnarray*}
(Cramer's Rule should be explained in your textbook, but you can also use any other method you like. The point is that you need to solve the system of linear equations given above.)


Notice that, in order to construct the partial derivatives of $x(u,v)$ and $y(u,v)$, we need the matrix of partial derivatives
$$
\begin{pmatrix}
f_1 & f_2 \\
g_1 & g_2
\end{pmatrix}
$$
to be invertible. Equivalently, the determinant of this matrix (which we call the \underline{Jacobian}) and write
$$
\frac{\partial (f,g)}{\partial (x,y)} = \frac{\partial (u,v)}{\partial (x,y)} = \begin{vmatrix}
f_1 & f_2 \\
g_1 & g_2
\end{vmatrix}
$$
must be non-zero. We see examples in the Worked Problem Section below.



\subsection{Differentiation in Higher Dimensions}
This approach for determining implicit functions may seem a bit \textit{ad hoc}, meaning that we seemingly worked backward from the conclusion in some cases.  In this section, we seek to remedy this a bit by providing a common framework for Questions 1 and 2. We will then sketch how our implicit differentiation formulas could be justified rigorously. 


Our common framework is based on \underline{vector-valued} functions $\mathbf{f} : \mathbb{R}^n \rightarrow \mathbb{R}^m$. (Instead of a number, the output of the function is an $m$-dimensional vector.) For short, we often write $\mathbf{f}(\mathbf{x}) = \mathbf{y}$ where $\mathbf{x} = \langle x_1,...,x_n \rangle$, $\mathbf{y} = \langle y_1,...,y_m\rangle$, and each $y_j$ satisfies
$$
y_j = f_j (x_1,...,x_n).
$$
Note: this is one situation where subscript notation does \underline{not} indicate differentiation. Rather,  $f_j$ represents the $j$-th component function of $\mathbf{f} = \langle f_1,...,f_m \rangle$. 

Our concepts of differentiability and linear approximation extend to this setting, as long as we remember that the linear approximation also needs to be vector-valued. 
Specifically, we want to use the following linear approximation near $\mathbf{x} = \mathbf{a}$,
$$
\mathbf{f} (\mathbf{x}) \approx \mathbf{L}(\mathbf{x}) := \mathbf{f} (\mathbf{a}) + D \mathbf{f} (\mathbf{a}) (\mathbf{x} - \mathbf{a}).
$$
Here, $D \mathbf{f}(\mathbf{a})$ denotes the \textbf{Derivative Matrix} (sometimes also called the \textbf{Jacobian Matrix}) of $\mathbf{f}$, defined as
$$
D \mathbf{f}(\mathbf{a})
=
\begin{pmatrix}
\frac{\partial y_1}{\partial x_1} & \cdots & \frac{\partial y_1}{\partial x_n} \\
\vdots & \quad & \vdots \\
\frac{\partial y_m}{\partial x_1} & \quad & \frac{\partial y_m}{\partial x_n} 
\end{pmatrix},
$$
where all partial derivatives are evaluated at $\mathbf{x}=\mathbf{a}$.  
We say that $\mathbf{f}$ is \underline{differentiable} at $\mathbf{a}$ if $\mathbf{R}(\mathbf{x}) = \mathbf{f}(\mathbf{x}) - \mathbf{L}(\mathbf{x})$ satisfies
$$
\lim_{\mathbf{x} \rightarrow \mathbf{a}} \frac{|\mathbf{R}(\mathbf{x})|}{|\mathbf{x} - \mathbf{a}|} = 0.
$$
We also have a vector-valued version of the Chain Rule: 
if $\mathbf{f} : \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $\mathbf{g} : \mathbb{R}^m \rightarrow \mathbb{R}^l$ are vector-valued functions whose all components are $C^1$, then,
$$
D(\mathbf{g} \circ \mathbf{f}) (\mathbf{x}) = \underbrace{D\mathbf{g}(\mathbf{f}(\mathbf{x}))}_{l \times m \textrm{ matrix}} \underbrace{D\mathbf{f}(\mathbf{x})}_{m \times n \textrm{ matrix}}.
$$
Notice that,  in the vector-valued case, we replace scalar multiplication with matrix multiplication.

The general form of the Implicit Function Theorem is, then, as follows. 

\begin{theorem}
Let ${\bf F}({\bf x},{\bf y})$ be a vector-valued function from a set $S\subset \RR^{n+k}$ to $\RR^k$, where ${\bf x}=(x_1,\dots,x_n)$ and ${\bf y}=(y_1,\dots,y_k)$. Assume that:
\begin{itemize}
\item ${\bf F}({\bf a},{\bf b})=0$ for some $({\bf a},{\bf b})\in S$,
\item ${\bf F}$ is $C^1$ in a neighbourhood of $({\bf a},{\bf b})$, meaning that its components $F_1,\dots,F_k$ are all $C^1$ in that neighbourhood, 
\item the $k\times k$ matrix
$$
D_{{\bf y}} \mathbf{F}(\mathbf{a},{\bf b})
=
\begin{pmatrix}
\frac{\partial F_1}{\partial y_1} & \cdots & \frac{\partial F_1}{\partial y_k} \\
\vdots & \quad & \vdots \\
\frac{\partial F_k}{\partial y_1} & \quad & \frac{\partial F_k}{\partial y_k} 
\end{pmatrix}
$$
is invertible. 
\end{itemize}
Then there is a (possibly smaller) neighbourhood of $({\bf a},{\bf b})$ where we can solve the equation
${\bf F}({\bf x},{\bf y})=0$
for ${\bf y}$ as a $C^1$ vector-valued function of ${\bf x}$. Moreover, the derivatives $\partial y_i/\partial x_j$ can be found by differentiating implicitly, and then solving, the system of equations ${\bf F}({\bf x},{\bf y}({\bf x}))=0$.
\end{theorem}

The proof of the theorem is a bit technical (see your textbook for more details), but the general idea is to use linear approximation. If the function ${\bf F}$ were actually linear, we could just set up a system of linear equations for ${\bf y}$ and then solve it. The difficulty is in extending the argument to functions that are $C^1$: they are no longer linear, but they are still well approximated by linear functions near every point, and those linear functions do not change too much as we move from one point to another.


We now return to our Questions 1 and 2, with a view to using linear approximation to justify the calculations we did.

\medskip
\noindent
\underline{Question 1:} First,  suppose again that $F(x,y,z) = 0$ is the equation relating the variables $x$, $y$ and $z$. (This is just a real-valued function, so vector-valued functions are not needed for this part.) We first solve the case when the function $F$ is linear in its variables, so that our equation has the form
$$
F(x,y,z) = Ax + By + Cz + D = 0.
$$
Assuming $C \neq 0$,  we solve for $z$ to find
$$
z = \frac{1}{C} (-Ax - By - D) \Longrightarrow \frac{\partial z}{\partial x} = - \frac{A}{C}, \,  \, \, \frac{\partial z}{\partial y} = - \frac{B}{C}.
$$
However, this corresponds to equations \eqref{eq:34eq2} and \eqref{eq:34eq3}! Why so? Well,  when $F(x,y,z) = Ax + By + Cz + D$ we have
$$
F_1 = A, F_2 = B, F_3 = C \ \Rightarrow \ \frac{\partial z}{\partial x} = - \frac{F_1}{F_3} \textrm{ and } \frac{\partial z}{\partial y} = - \frac{F_2}{F_3}.
$$
For general $C^1$ functions $F(x,y,z)$---that is, functions which are not assumed to be linear---we may instead approximate by a linear function $L(x,y,z)$.  The condition that $F(x,y,z)$ is $C^1$ guarantees that this linearization works as a suitable approximation and that the approximation errors do not spoil the picture. 

\medskip
\noindent
\underline{Question 2:} For systems of equations, we write
$$
\mathbf{F}(x,y) =
\begin{pmatrix}
f(x,y) \\
g(x,y)
\end{pmatrix}
$$
which is a vector-valued function.  The assumption that $f(x,y)$ and $g(x,y)$ are $C^1$ allows us to use the linear approximation
$$
\mathbf{F}(x,y) \approx \mathbf{F}(a,b) + D \mathbf{F}(a,b) \begin{pmatrix} x - a \\ y - b \end{pmatrix},  \textrm{ where } 
D \mathbf{F} : =
\begin{pmatrix}
f_1 & f_2 \\
g_1 & g_2
\end{pmatrix}
$$
If, in this situation, the functions $f(x,y)$ and $g(x,y)$ were linear in their variables, we could simply solve the linear system
$$
\begin{cases}
u = f(x,y) \\
v = g(x,y)
\end{cases}
$$
for $x$ and $y$, then differentiate the corresponding solution. A formal proof of the Implicit Function Theorem uses this linearization as an approximation  and sets up a linear system of equations for the derivatives.  The crux of the proof, then,  is proving that the error introduced by this linearization is controllable---a fact which, again, is guaranteed by the assumption that $f$ and $g$ are $C^1$. 

\medskip




%%%%%%%%%%%%%%%%%%%%


\subsection{Implicitly defined curves and surfaces} This is a good time to return to the question of when an equation $f(x,y)=c$ defines a curve in the plane, or when an equation $f(x,y,z)=C$ defines a surface in $\RR^3$. 

We will look at the equation $f(x,y)=c$ first.  Let $F(x,y)=f(x,y)-c$, so that the equation $f(x,y)=c$ is equivalent to $F(x,y)=0$. Assume that $f$ is $C^1$ on some neighbourhood of $(x,y)=(a,b)$, and that $\nabla f (a,b) \neq {\bf 0}$. Then at least one of the partials $f_1(a,b),f_2(a,b)$ is non-zero. Assume that $f_2(a,b)\neq 0$. Then also $F_2(a,b)\neq 0$, and we can apply the Implicit Function Theorem to say that there is a neighbourhood of $(a,b)$ where we can solve for $y=y(x)$ as a $C^1$ function. This means that, in that neighbourhood, the graph $f(x,y)=c$ \underline{is a curve} with a $C^1$ parametrization $(x,y(x))$. If we assume that $f_1(a,b)\neq 0$ instead, then the same is true with $x$ and $y$ interchanged.

For equations $f(x,y,z)=C$ in 3 variables, the situation is similar. Assume that $f$ is $C^1$ on some neighbourhood of $(x,y,z)=(a,b,c)$, and that $\nabla f (a,b,c) \neq {\bf 0}$. Without loss of generality, we may assume that $f_3(a,b,c)\neq 0$. Then, again by the Implicit Function Theorem, the equation $f(x,y,z)=C$ has a $C^1$ solution $z=z(x,y)$ in some neighbourhood of $(a,b,c)$, describing a surface with a nice parametrization.

This is why the $C^1$ assumption in Theorem  \ref{thm:33thm1} implies that the level set $f({\bf x})=f({\bf a})$ is, at least locally, a surface with a $C^1$ parametrization (a curve if $n=2$). If we knew this up front (for some other reason), then it would be enough to assume that $f$ is just differentiable at ${\bf a}$, and not necessarily $C^1$ on its neighbourhood.



%%%%%%%%%%%%%%%%

\section*{Worked Problems}
\begin{enumerate}[1.]
\item {\it A certain quadric surface is given by the equation $y^2 = 2x^2 + z^2 - 1$.  Find the value of $\frac{\partial z}{\partial x}$ at the point $(2,4,3)$ on the surface.}

\medskip
The surface is given by the equation $F(x,y,z) = 0$, where $F(x,y,z)$ is the function $F(x,y,z) = 2x^2 - y^2 + z^2 - 1$. Differentiating implicitly in $x$ gives
$$
\frac{\partial z}{\partial x}\vert_{(x,y,z) = (2,4,3)} = - \frac{F_x}{F_z}\Big\vert_{(x,y,z) = (2,4,3)} = - \frac{4x}{2z}\Big\vert_{(x,y,z) = (2,4,3)} = - \frac{4}{3}.
$$

\medskip
In this question, an alternative method would be to solve for the variable $z$ \underline{explicitly}, so that
$$
z^2 = y^2 - 2x^2 + 1 \ \Rightarrow \ z = \sqrt{y^2 - 2x^2 + 1},
$$
and then differentiate in $x$. (Note: we chose the $+$ square-root because at $(2,4,3)$, we know that $z = 3 > 0$.) This method is not available in Question 2 below.




\begin{figure}[H]
\centering
\includegraphics[width= .5\textwidth]{34fig2.png}
\caption{The surface given by the equation $F(x,y,z) = 0$. }
\end{figure}



\medskip

\item {\it Suppose $x,y$ and $z$ satisfy the equation
$
x + x \sin y + y \sin z = 2 z \cos x.
$
Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at  the point $(0, \pi, \frac{\pi}{2})$.}

\medskip

This time, we cannot really solve for $z$ in any explicit form, so that implicit differentiation is the only method we can use.
First differentiate the equation in $x$, under the assumption that $z : = z(x,y)$, which gives
$$
\frac{\partial}{\partial x}(x + x \sin y + y \sin z) = \frac{\partial}{\partial x}(2 z \cos x)
$$
so that
$$
1 + \sin y + y \cos z \cdot \frac{\partial z}{\partial x} = 2 \frac{\partial z}{\partial x} \cos x - 2 z \sin x.
$$
Now use that $(x,y,z) = (0, \pi, \frac{\pi}{2})$ to obtain
$$
1 = 2 \frac{\partial z}{\partial x} \Longrightarrow \frac{\partial z}{\partial x} = \frac{1}{2}.
$$
A similar differentiation in $y$ gives
$$
0 + x \cos y + \sin z + y \cos z \frac{\partial z}{\partial y} = 2 \frac{\partial z}{\partial y} \cos x.
$$
Evaluating at $(x,y,z) = (0, \pi, \frac{\pi}{2})$ gives $\frac{\partial z}{\partial y} = \frac{1}{2}$.


\bigskip

\item {\it Suppose we have the system of equations
$$
\begin{cases}
u = 2x^2 + y \\
v = y^2 - 4x
\end{cases}
$$
which implicitly defines $x$ and $y$ as functions of $u$ and $v$. Find $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial u}$ wherever possible.
}

\medskip

Differentiate the system of equations in $u$, keeping in mind that $x = x(u,v)$ and $y = y(u,v)$, and applying the Chain Rule accordingly. This gives
\begin{eqnarray*}
1 & = & \frac{\partial u}{\partial u} = \frac{\partial}{\partial u} (2x^2 + y) = 4x \frac{\partial x}{\partial u} +  \frac{\partial y}{\partial u} \\
0 & = & \frac{\partial v}{\partial u} = \frac{\partial}{\partial u} (y^2 - 4x) = 2y \frac{\partial y}{\partial u} -4 \frac{\partial x}{\partial u}
\end{eqnarray*}
We simplify the second equation to $0= -2 \frac{\partial x}{\partial u} +y \frac{\partial y}{\partial u}$,
and then solve using Cramer's rule:
\begin{eqnarray*}
\frac{\partial x}{\partial u} & = & \frac{\begin{vmatrix} 1 & 1 \\ 0 & y \end{vmatrix}}{\begin{vmatrix} 4x & 1 \\ -2 & y \end{vmatrix}} = \frac{y}{4xy + 2}  \\[2ex]
\frac{\partial y}{\partial u} & = & \frac{\begin{vmatrix} 4x & 1 \\ -2 & 0 \end{vmatrix}}{\begin{vmatrix} 4x & 1 \\ -2 & y \end{vmatrix}} = \frac{2}{4xy + 2} = \frac{1}{2xy + 1} 
\end{eqnarray*}
Notice that both formulae are valid so long as $2xy + 1 \neq 0$, so that $xy \neq -1/2$ suffices.
\end{enumerate}


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

\bigskip

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

\bigskip

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

\bigskip


\item {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.}

\bigskip


\item
{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 $(x,y)=(1,0)$.}


\bigskip

\item {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
$x=-1$, $y=1$.}

\bigskip

\item
{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: $\sin(\pi/3)=\frac{\sqrt{3}}{2}$,
$\cos(\pi/3)=\frac{1}{2}$.)}

\end{enumerate}





