# Section 2.6

## Table of Contents

Implicit Differentiation

# Implicit curves #

Some curves are defined *implicitly* as equations. They’re not necessarily functions, but we can still do calculus.^{1}

## Example: #

The equation $x^2 + y^2 = 4$ defines a curve - a circle!

It consists of two *implicit* functions $y = \sqrt{4-x^2}$ and $y = - \sqrt{4-x^2}$, the upper and lower arcs of the circle, respectively.

# Implicit differentiate #

- Our process is to take the derivative of both sides of the equation with respect to $x$.
- When you differentiate a $y$, we use the chain rule and multiply by $\frac{dy}{dx}$.

## Example #

Differentiate $x^2 + y^2 = 4$ implicitly and solve for $\frac{dy}{dx}$, then find the points where $x^2 + y^2 = 4$ where the tangent line is horizontal and where the tangent line is vertical.

## Example #

Find $\dfrac{dy}{dx}$ implicitly if $x\sin y + y\sin x = 1$

This is a really neat wobbly curve:

Here I’m using Geogebra to model the tangent line to the curve at any arbitrary point. Pretend that the curve models the shoreline; the tangent line would tell our GPS exactly how to hug the shore without crashing.

Check out this geogebra activity to play with it: (click this link)

https://www.geogebra.org/calculator/nzuk449p

## The second derivative #

In order to find the second derivative with respect to $x$, we first need to solve for $\dfrac{dy}{dx}$ and differentiate that quantity.

The notation should help you remember this: $$ \dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\frac{dy}{dx} $$

### Example: #

Find $y’’$ of the curve $x^3 + y^3 = 1$

### Example #

Find the equation of the of the tangent line at the point $(0, 1/2)$ of the curve $$ x^2 + y^2 = (2x^2 + 2y^2 - x)^2 $$

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