
# Implicit Differentiation

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

## Implicit and Explicit Functions

An explicit function is an function expressed as y = f(x) such as

$y = \text{sin}\; x$

y is defined implicitly if both x and y occur on the same side of the equation such as

$x^2 + y^2 = 4$

we can think of y as function of x and write:

$x^2 + y(x)^2 = 4$

## Implicit Differentiation

To find dy/dx, we proceed as follows:

1. Take d/dx of both sides of the equation remembering to multiply by y' each time you see a y term.
2. Solve for y'

Example
Find dy/dx implicitly for the circle

$x^2 + y^2 = 4$

Solution

1. d/dx (x2 + y2) = d/dx (4)

or

2x + 2yy' = 0

2. Solving for y, we get

2yy' = -2x

y' = -2x/2y

y' = -x/y

Example:

Find y' at (4,2) if

$xy + \dfrac{x}{y} = 10$

Solution:

1. $(xy)' + \left(\dfrac{x}{y}\right)' = (5)'$
Using the product rule and the quotient rule we have

2. $xy' + y + \dfrac{y - xy'}{ y^2} = 0$

3. Now plugging in x = 4 and y = 2,

2 - 4y'
4y' + 2 + = 0
22

16y' + 8 + 2 - 4y' = 0 Multiply both sides by 4

12y' + 10 = 0

12y' = -10

y' = -5/6

## Exercises

1. Let
$3x^2 - y^3 = 4x \text{cos}\; x + y^2$
Find dy/dx

2. Find dy/dx at (-1,1) if
$(x + y)^3 = x^3 + y^3$

3. Find dy/dx if
$x^2 + 3xy + y^2 = 1$

4. Find y'' if
$x^2 - y^2 = 4$

Larry Green (Lake Tahoe Community College)