Implicit Functions & Differentiation

1. Consider the circle defined by \( x^2 + y^2 = 25\) and the tangent line touching the circle at \((3,4)\).
   What is the slope of this tangent?
2. What is the slope of the radius passing through \((3,4)\)?
3. How are these related?

1. Consider the circle defined by \( x^2 + y^2 = 25\) and the tangent line touching the circle at \((3,4)\).
   What is the slope of this tangent? \( -\frac{3}{4}\)
2. What is the slope of the radius passing through (3,4)? \( \frac{4}{3}\)
3. How are these related? The lines are perpendicular (or the slopes multiply to \(-1\))

A 'mapping' from one set to another such that each valid 'in' is mapped to only one 'out'

What functions do you know already?

1. \(x, x + 3, x^2, \sqrt{x}\)
2. \(\sin(x), \cos(x), \tan(x)\)
3. \(\mathbb{E}, \frac{d}{dx}\)

Is integration a function?

So what can be said about the graphs of functions?

Let's restrict our attention to graphable functions, that is, to functions from \(\mathbb{R}\) to \(\mathbb{R}\).

An unstated assumption of what we said before was that we were looking at functions like \(y = x^2\) or \(y = \cos(3x)\) where the \(y\) was the subject of the equation.

This made it very clear what was the 'in' (\(x\)) and what was the 'out' (\(y\))

But what does it imply about the graphs?

Vertical Line Test

The graph of a function will never have multiple points of intersection with a vertical line.

Horizontal Line Test

If the graph of a function never crosses a horizontal line multiple times, then it is one-to-one and its inverse is a function.

The preceeding is somewhat restrictive: there are lots of nice curves which don't pass the vertical line test.

The preceeding is somewhat restrictive: there are lots of nice curves which don't pass the vertical line test.

But we like gradients! And we like the ease of use derivatives provide us.

What can we do? When in doubt, cheat!

Implicit functions are a means of defining a 'function' (or something close enough) by writing an equation in terms of \(x\) and \(y\) without making \(y\) the subject.

\[4x^2 + y^2 = 36\]

\[\cos x + \cos y = \frac{1}{2}\]

Implicit Functions

One way to think about what implicit functions do is they define equations which could theoretically be solved for \(x\) and simply assert that they are 'good enough' for defining a function.

If there are issues like \(y = \dots \pm \sqrt{\dots}\) we just fuhgeddaboudit.

These are no longer functions in the strict sense because they don't pass the vertical line test: rather, we implicitly choose the relevant 'branch cut' and assume we are locally dealing with a function.

So how does this all tie back to gradients? Let's think about the chain rule: \[\begin{align*} \frac{d}{dx}\sin(x^2) & = \sin'(x^2) * (x^2)' \\ & = 2x \cos(x^2) \\ \frac{d}{dx}f(x^2) & = f'(x^2) * (x^2)' \\ & = 2x * f'(x^2) \\ \frac{d}{dx} f(x) & = f'(x) \end{align*}\]

\(y\) is no more or less a function of \(x\) than \(f\) is here. \[\frac{d}{dx}y(x) = y'(x)\] or simply \[\frac{d}{dx}y = y'\]

\(y\) can be differentiated in complex contexts much the same way, by smart application of the usual chain and product rules: \[\begin{align*} \frac{d}{dx} \sqrt{xy^2} & = \frac{d}{dx}\left(x^{1/2}y\right) \\ & = \frac{1}{2}x^{1/2 - 1} * y + x^{1/2}\frac{d}{dx}(y) \\ & = \frac{1}{2}x^{-1/2}y + x^{1/2}y' \\ & = \frac{y}{2\sqrt{x}} + y'\sqrt{x} \end{align*}\]

Or even like this by applying the chain rule repeatedly: \[\begin{align*} \frac{d}{dx} \sqrt{xy^3} & = \frac{d}{dx}\left(x^{1/2}y^{3/2}\right) \\ & = \frac{1}{2}x^{1/2 - 1} * y^{3/2} + x^{1/2}\frac{d}{dx}(y^{3/2}) \\ & = \frac{1}{2}x^{-1/2}y + x^{1/2}\left(\frac{3}{2}y^{3/2 - 1}*y'\right) \\ & = \frac{y}{2\sqrt{x}} + \frac{3}{2}y'\sqrt{xy} \end{align*}\]

A key property of functions is that they always send the same input to the same output. This is why we can apply the same function to two sides of an equation and get another equation back. Notably, this holds for the \(\frac{d}{dx}\) function:

If \(f(x) = g(x)\) then \(f'(x) = g'(x)\).

Or if \(f(x,y) = g(x,y)\) then \(\frac{d}{dx}f(x,y) = \frac{d}{dx}g(x,y)\).

A common (and useful!) trick is to take logarithms before differentiating: \(xy = 1\)

The long(-ish) way: \[\begin{align*} xy & = 1 \\ \frac{d}{dx}\left(xy\right) & = \frac{d}{dx}(1) \\ \frac{d}{dx}(x) * y + x * \frac{d}{dx}(y) & = 0 \\ 1 * y + x * y' & = 0 \\ y + xy' & = 0 \\ y' & = -\frac{y}{x} \end{align*}\]

Arguably easier: \[\begin{align*} xy & = 1 \\ \ln x + \ln y & = 0 \\ \frac{1}{x} + \frac{y'}{y} & = 0 \\ y' & = - \frac{y}{x} \end{align*}\]

A harder example: \[\begin{align*} x^2y^3 & = 1 \\ \frac{d}{dx}\left(x^2y^3\right)& = 0 \\ 2xy^3+ x^2 * 3y^2 * y' & = 0\\ y' & = - \frac{2xy^3}{3x^2y^2} \\ & = -\frac{2y}{3x} \end{align*}\]

Or ... \[\begin{align*} x^2y^3 & = 1 \\ 2\ln(x) + 3\ln(y) & = 0 \\ \frac{2}{x} + \frac{3y'}{y} & = 0 \\ \frac{3y'}{y} & = - \frac{2}{x} \\ y' & = -\frac{2y}{3x} \end{align*}\]

Consider the circle defined by \( x^2 + y^2 = 25\) and the tangent line touching the circle at \((3,4)\).
What is the slope of this tangent?

\[\begin{align*} x^2 + y^2 & = 25\\ 2x + 2y * y' & = 0 \\ y' & = -\frac{x}{y}\\ \implies y' & = -\frac{3}{4} \text{ at } (3,4) \end{align*}\]