- 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 signfor 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 changeswith 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 functionof 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 linearfunction. $y = f(p + \delta) = f(p) + f'(p) \delta$. Now since $f'(p)$ islocally 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 generalizesin 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 $x[n+1] = x[n] - (y^* - y[n]) / f'(x[n])$.