- Example 1: $x^2 + y^2 = 1$ to $y = \sqrt{1 - x^2}$.

- Note that the solution exists iff $\alpha \neq 0$.
- Also note that the the existence of the solution is equivalent to asking that $\partial h / \partial y = \alpha \neq 0$.

- We can however build two functions by taking two parts. $y+ = +\sqrt{p^2 - 1}$; $y- = -\sqrt{p^2 - 1}$.

- In this case, we have $\partial h / \partial y = 2y$, which changes sign for the two solutions. If $y^\star > 0$, then $(\partial h / \partial y)(y^\star = 0)$. Similarly for the negative case.

$\begin{aligned}
&3 sol(p)^2 sol'(p) + 2p - 3 sol'(p) p - 3 sol(p) = 0 \\
&sol'(p)\left[ 3 sol(p)^2 - 3p \right] = 3 sol(p) - 2p \\
&sol'(p) = [3 sol(p) - 2p] / \left[ 3(sol(p)^2 - 3p) \right] \\
\end{aligned}$

The above solution exists if $3(sol(p)^2 - 3p \neq 0)$. This quantity is again
$\partial h / \partial y$.
- We have two inputs which are purchaed as $x_1$ units of input 1, $x_2$ units of input $2$.
- The price of the first input is $w_1 BTC/unit$. That of the second input is $w_2 BTC/unit$.
- We produce an output which is sold at price $w BTC/unit$.
- For a given $(x_1, x_2)$ units of input, we can produce $x_1^a x_2^b$ units of output where $a + b < 1$. The Coob-douglas function .
- The profit is going to be $profit(x_1 x_2, w_1, w_2, w) = w(x_1^a x_2^b) - w_1 x_1 - w_2 x_2$.
- We want to select $x_1, x_2$ to maximize profits.
- Assume we are at break-even: $profit(x_1, x_2, w_1, w_2, w) = 0$.
- The implicit function theorem allows us to understand how any variable changes with respect to any other variable. It tells us that locally, for example, that the number of units of the first input we buy ( $x_1$) is a function of the price $w_1$. Moreover, we can show that it's a
*decreasing function*of the price.

- Given a differentiable function $f$, at a point $p$, we will have a continuous inverse $f^{-1}(p)$ if the derivative $f'(p)$ is locally invertible.

- The intuition is that we can approximate the original function with a linear function. $y = f(p + \delta) = f(p) + f'(p) \delta$. Now since $f'(p)$ is locally invertible, we can solve for $\delta$. $y = f(p) + f'(p) \delta$implies that $\delta = 1/f'(p) [y - f(p + \delta) ]$. This gives us the pre-image $(p + delta) \mapsto y$.

- The fact that $1/f'(p)$ is non-zero is the key property. This generalizes in multiple dimensions to saying that $f'(p)$ is invertible.

- We start with some $x[1]$.
- We then find the tangent $f'(x[1])$.
- We draw the tangent at the point $x[1]$ as $(y[2] - y[1]) = f'(x[1])(x[2] - x[1])$.
- To find the $y^*$ we set $y[2] = y^*$.
- This gives us $x[2] = x[1] - (y^* - y[1])/f'(x[1])$.
- Immediately generalizing, we get $x[n+1] = x[n] - (y^* - y[n]) / f'(x[n])$.

- Let $F(x, y) \neq 0$, point $p$ be the point where we wish to implicitize.
- To apply implicit fn theorem, take $\partial_y|_pF(x, y) \neq 0$. Say WLOG that $\partial_y|_p F(x, y) > 0$, since $F$ is assumed to be continuously differentiable.
- Since $\partial_y F$ is continuous and positive at $p$, it's positive in a nbhd of $p$ by continuity.
- Consider $F(p_x, y)$ as a single variable function of $y$. Its derivative with respect to $y$ is positive; it's an increasing function in terms of $y$.
- Since $F(x_0, y_0)$ and is an increasing function of $y_0$, we must have two $y$ values $y_-, y_+$ such that $F(p_x, y_+)$ is positive, and $F(p_x, y_-)$is negative.
- Since $F$ is zero and continuous, we have that for all $x$ near $x_0$, that $F(x_0, y_+) > 0 \implies F(x, y_+) > 0$ and $F(x_0, y_-) < 0 \implies F(x, y_-) < 0$. We have released $x_0$ into a wild $x$ in the neighbourhood!
- Now pick some $x_*$ near $x$. Since we have that $F(x_*, y_+) > 0$ and $F(x_*, y_-) < 0$, there exists a
$y_*$ (by MEAN VALUE THEOREM) that $F(x_*, y_*) = 0$*unique* - Since the $y_*$ is unique, we found a function: $x_* \xmapsto{f} y_*$.
- We are not done. We need to prove the formula for $f'(x)$ where $f$ is the implicit mapping.
- $F(x, f(x)) = 0$ by defn of $f(x)$. Apply chain rule!

$\begin{aligned}
&F(x, f(x)) = 0 \\
&dF = 0 \\
&\frac{\partial F}{\partial x} \cdot \frac{\partial x}{\partial x} + \frac{\partial F}{\partial y} \frac{df}{dx} = 0 \\
&\frac{\partial F}{\partial x} \cdot 1 + \frac{\partial F}{\partial y} \frac{df}{dx} = 0 \\
&\frac{df}{dx} = \frac{- \frac{\partial F}{\partial y}}{\frac{\partial F}{\partial x} \cdot 1} \\
\end{aligned}$

- Given the implicit function, say we want to locally invert $f(x)$.
- Pick the implicit function $F(x, y) = f(y) - x$. If we consider the level set $F(x, y) = 0$, the implicit function theorem grants us a $g(x)$ such that $F(x, y=g(x)) = 0$. That is, we get $f(g(x)) - x = 0$, or $f(g(x)) = x$.
- To show that it's also a right inverse, consider $F(f(k), k) = f(k) - f(k) = 0$. Since $y = g(x)$, we have that $k = g(f(k))$.
- Hence, $g$ and $f$ are both left and right inverses, and hence bijections.

- We are given a function $F(x, y)$. We wish to find a formula $y = g(x)$ such that $F(x, g(x)) = 0$.
- The idea is to consider a function $f(x, y) = (x, F(x, y))$.
- Since this is invertible, we get a local inverse function $g$ such that $g(f(x, y)) = (x, y)$. That is, $g(x, F(x,y)) = (x, y)$.
- Now, for a given $x$, set $y := snd(g(x, 0))$. This gives us a $(x, y)$ such that $g(x, F(x,y)=0) = (x, y)$. That is, we get a $y$ such that $F(x, y) = 0$ and this a bijection from $x$ to $y$ since $g$ is a bijection.