Implict differentiation

Section 4.6 Implict differentiation

So far we have mostly dealt by functions that are given “explicitly”, as

y = f (x) .

$y = f(x)\text{.}$ But sometimes functions are defined “implicitly”: this happens when a function

y = f (x)

$y = f(x)$ is defined implicitly by a relation between

y

$y$ and

x

$x$ (which you may or may not be able to solve explicitly for

y

$y$ ). We study such implicit functions in this section, and show how we can calculate their derivatives. The resulting process is known as “implicit differentiation”.

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Objectives

You should be able to:

Explain and illustrate the concept of implicit functions.
Use the method of implicit differentiation to calculate the derivative of implicit functions.

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Subsection 4.6.1 Instructional video

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Subsection 4.6.2 Key concepts

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Concept 4.6.1. Implicit functions.

Given a relation

H (x, y) = 0,

$\begin{equation*} H(x,y)=0, \end{equation*}$

we say that $f$ is a function defined implicitly by this relation if

H (x, f (x)) = 0

$\begin{equation*} H(x, f(x) ) = 0 \end{equation*}$

for all $x$ in the domain of $f\text{.}$

A relation $H(x,y)=0$ defines a curve in the $xy$ -plane, which implicitly defines $y$ as one or several functions of $x\text{.}$

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Concept 4.6.2. Implicit differentiation.

Implicit differentiation is the process of calculating the derivative of a function defined implicitly by a relation $H(x,y)=0\text{.}$

To calculate the derivative $y'$ of an implicit function:

Treat the variable $y$ in the relation as an unknown but differentiable function of $x$ (like $y=g(x)$ ), and differentiate both sides of the relation with respect to $x\text{,}$ using the chain rule.
Collect the terms involving $y'$ on one side of the equation and solve for $y'\text{.}$

Note that this will generally give $y'$ as a function of $x$ and $y\text{,}$ where $y$ is understood as the function implicitly defined by the original relation.

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MATH 144: Calculus for the Physical Sciences I