Implicit Differentiation

Finding slopes on curves you can't (or don't want to) solve for y.

0 / 10 problems mastered

A curve like $x^3+y^3=6xy$ isn't a function $y=f(x)$ , but near most points it looks like one. So treat $y$ as an unknown function of $x$ and differentiate both sides with respect to $x$ .

The one rule: every time you differentiate something with $y$ in it, the chain rule attaches a factor of $y'$ . So $\frac{d}{dx}(y^3)=3y^2\,y'$ , $\frac{d}{dx}(\sin y)=\cos y\,y'$ , and $\frac{d}{dx}(xy)=y+x\,y'$ (product rule).

Recipe:

Differentiate both sides with respect to $x$ . Attach $y'$ to every $y$ -term.
Move all $y'$ terms to one side, everything else to the other.
Factor out $y'$ and divide.

Tangent lines on implicit curves

Lab 7 · drag the point along the curve

curve

point (x, y)–

dy/dx formula–

slope here–

When the denominator of dy/dx is 0, the tangent is vertical. When the numerator is 0, it is horizontal.

Worked examples

Example · folium of Descartes (Stewart Ex. 2)

Find $y'$ for $x^3+y^3=6xy$ and the tangent line at $(3,3)$ .

Differentiate: $3x^2+3y^2y'=6y+6xy'$ . Right side used the product rule on $6xy$ .
Collect $y'$ : $3y^2y'-6xy'=6y-3x^2$ .
Factor and divide (and cancel 3): $y'=\dfrac{2y-x^2}{y^2-2x}$ .
At $(3,3)$ : $y'=\dfrac{6-9}{9-6}=-1$ . Tangent: $y-3=-(x-3)\Rightarrow x+y=6$ .

Example · second derivative (§3.5 #39)

Find $y''$ for $x^2+4y^2=4$ .

$2x+8yy'=0\Rightarrow y'=-\dfrac{x}{4y}$ .
Quotient rule on $y'$ , remembering $y$ is a function: $y''=-\dfrac{4y\cdot1-x\cdot4y'}{16y^2}$ .
Substitute $y'=-\frac{x}{4y}$ : $y''=-\dfrac{4y+\frac{x^2}{y}}{16y^2}=-\dfrac{4y^2+x^2}{16y^3}$ .
Use the original equation: $x^2+4y^2=4$ . $y''=-\dfrac{4}{16y^3}=-\dfrac{1}{4y^3}$ .

Practice

§3.5 #1

Find

y'

for

5x^2-y^3=7

implicitly.

§3.5 #5

Find

dy/dx

for

x^2-4xy+y^2=4

§3.5 #7

Find

dy/dx

for

x^4+x^2y^2+y^3=5

§3.5 #10

Find

dy/dx

for

xe^y=x-y

§3.5 #11

Find

dy/dx

for

\sin x+\cos y=2x-3y

§3.5 #15

Find

dy/dx

for

y\cos x=x^2+y^2

§3.5 #23

f(x)+x^2[f(x)]^3=10

and

f(1)=2

, find

f'(1)

§3.5 #37

Find the tangent line to the kampyle

y^2=5x^4-x^2

(1,2)

§3.5 · circle

Find the tangent line to

x^2+y^2=25

(3,4)

§3.5 · horizontal tangents

Where does

x^2-xy+y^2=3

have horizontal tangents?

Traps

Forgetting y′ on y-terms

$\frac{d}{dx}y^2=2y\,y'$ , not $2y$ . This is the #1 mistake.

xy is a product

$\frac{d}{dx}(xy)=y+xy'$ . Don't write just $y'$ or $1\cdot y'$ .

Constant on the right

The derivative of the 25 in $x^2+y^2=25$ is 0. Don't drop the whole right side when it has variables.