Lesson 16: Implicit Differentiation

Lesson 16 (Section 3.6)Implicit Differentiation

Derivatives of Inverse Functions

Math 1a

October 31, 2007

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I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

ProblemFind the slope of the line which is tangent to the curve

x2 + y2 = 1

at the point (3/5,−4/5).

Solution (Old Way)

y2 = 1− x2

=⇒ y = −√

1− x2

dy

dx= − −2x

2√

1− x2=

x√1− x2

dy

dx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/5

4/5=

3

4.

ProblemFind the slope of the line which is tangent to the curve

x2 + y2 = 1

at the point (3/5,−4/5).

Solution (Old Way)

y2 = 1− x2

=⇒ y = −√

1− x2

dy

dx= − −2x

2√

1− x2=

x√1− x2

dy

dx

∣∣∣∣x=3/5

=3/5√

1− (3/5)2=

3/5

4/5=

3

4.

Solution (New Way)

Differentiating this, but pretend that y is a function of x. We have:

2x + 2ydy

dx= 0,

ordy

dx= −x

y.

So if x =3

5, y = −4

5, and r = 1, then the slope of the line tangent

to the circle isdy

dx

∣∣∣∣( 3

5,− 4

5 )=

3/5

4/5=

3

4.

Summary

any time a relation is given between x and y, we may differentiatey as a function of x even though it is not explicitly definedThis is called implicit differentiation.

Example

Find the equation of the line tangent to the curve

y2 = x2(x + 1) = x3 + x2

at the point (3,−6).

SolutionDifferentiating the expressionimplicitly with respect to x

gives 2ydy

dx= 3x2 + 2x , so

dy

dx=

3x2 + 2x

2y, and

dy

dx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −11

4.

Thus the equation of thetangent line is

y + 6 = −11

4(x − 3).

Example


y2 = x2(x + 1) = x3 + x2



gives 2ydy

dx= 3x2 + 2x , so

dy

dx=

3x2 + 2x

2y, and

dy

dx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −11

4.


y + 6 = −11

4(x − 3).

Example


y2 = x2(x + 1) = x3 + x2



gives 2ydy

dx= 3x2 + 2x , so

dy

dx=

3x2 + 2x

2y, and

dy

dx

∣∣∣∣(3,−6)

=3 · 32 + 2 · 3

2(−6)= −11

4.


y + 6 = −11

4(x − 3).

Example

Find the horizontal tangent lines to the same curve: y2 = x3 + x2

Solution

We need 3x2 + 1 = 0, or x =1√3

. Then

y2 =1

3√

3+

1

3,

Thus

y = ±

√1

3+

1

3√

3

Example

Find the horizontal tangent lines to the same curve: y2 = x3 + x2

Solution

We need 3x2 + 1 = 0, or x =1√3

. Then

y2 =1

3√

3+

1

3,

Thus

y = ±

√1

3+

1

3√

3

An example from chemistry

Example

The van der Waals equationdescribes nonideal propertiesof a gas:(

P + an2

V 2

)(V−nb) = nRT ,

where P is the pressure, Vthe volume, T thetemperature, n the number ofmoles of the gas, R aconstant, a is a measure ofattraction between particlesof the gas, and b a measureof particle size.

An example from chemistry

Example

The van der Waals equationdescribes nonideal propertiesof a gas:(

P + an2

V 2

)(V−nb) = nRT ,

where P is the pressure, Vthe volume, T thetemperature, n the number ofmoles of the gas, R aconstant, a is a measure ofattraction between particlesof the gas, and b a measureof particle size.

DefinitionThe isothermic compressibility of a fluid is defined by

β = −dV

dP

1

V

with temperature held constant.

The smaller the β, the “harder” the fluid.

Math 1a - October 31, 2007.GWBWednesday, Oct 31, 2007

Page5of13

DefinitionThe isothermic compressibility of a fluid is defined by

β = −dV

dP

1

V

with temperature held constant.

The smaller the β, the “harder” the fluid.

Let’s find the compressibility of a van der Waals gas.Differentiating the van der Waals equation by treating V as afunction of P gives(

P +an2

V 2

)dV

dP+ (V − bn)

(1− 2an2

V 3

dV

dP

)= 0,

so

β = − 1

V

dV

dP=

V 2(V − nb)

2abn3 − an2V + PV 3

I What if a = b = 0?

I What is the sign ofdβ

db?


da?


P +an2

V 2

)dV

dP+ (V − bn)

(1− 2an2

V 3

dV

dP

)= 0,

so

β = − 1

V

dV

dP=

V 2(V − nb)




db?


da?


P +an2

V 2

)dV

dP+ (V − bn)

(1− 2an2

V 3

dV

dP

)= 0,

so

β = − 1

V

dV

dP=

V 2(V − nb)




db?


da?


P +an2

V 2

)dV

dP+ (V − bn)

(1− 2an2

V 3

dV

dP

)= 0,

so

β = − 1

V

dV

dP=

V 2(V − nb)




db?


da?


P +an2

V 2

)dV

dP+ (V − bn)

(1− 2an2

V 3

dV

dP

)= 0,

so

β = − 1

V

dV

dP=

V 2(V − nb)




db?


da?


Page6of13

Theorem (The Inverse Function Theorem)

Let f be differentiable at a, and f ′(a) 6= 0. Then f −1 is defined ina neighborhood of b = f (a), and

(f −1)′(b) =1

f ′(f −1(b))

“Proof”.If y = f (x), take f −1 of both sides to get

x = f −1(f (x)),

so1 = (f −1)′(f (x))f ′(x).

Theorem (The Inverse Function Theorem)

Let f be differentiable at a, and f ′(a) 6= 0. Then f −1 is defined ina neighborhood of b = f (a), and

(f −1)′(b) =1

f ′(f −1(b))

“Proof”.If y = f (x), take f −1 of both sides to get

x = f −1(f (x)),

so1 = (f −1)′(f (x))f ′(x).

Using implicit differentiation to find derivatives

Example

Finddy

dxif y =

√x .

SolutionIf y =

√x, then

y2 = x ,

so

2ydy

dx= 1 =⇒ dy

dx=

1

2y=

1

2√

x.

Using implicit differentiation to find derivatives

Example

Finddy

dxif y =

√x .

SolutionIf y =

√x, then

y2 = x ,

so

2ydy

dx= 1 =⇒ dy

dx=

1

2y=

1

2√

x.

The power rule for rational numbers

Example

Finddy

dxif y = xp/q, where p and q are integers.

SolutionWe have

yq = xp

qyq−1 dy

dx= pxp−1

dy

dx=

p

q· xp−1

yq−1

Now yq−1 = xp(q−1)/q = xp−p/q so

xp−1

yq−1= xp−1−(p−p/q) = xp/q−1


Example

Finddy


SolutionWe have

yq = xp

qyq−1 dy

dx= pxp−1

dy

dx=

p

q· xp−1

yq−1


xp−1

yq−1= xp−1−(p−p/q) = xp/q−1


Example

Finddy


SolutionWe have

yq = xp

qyq−1 dy

dx= pxp−1

dy

dx=

p

q· xp−1

yq−1


xp−1

yq−1= xp−1−(p−p/q) = xp/q−1

Arcsin

Remember that

arcsin x = sin−1 x 6= 1

sin x

Example

Find the derivative of arcsin.

SolutionWe have y = arcsin x =⇒ x = sin y , so

1 = cos ydy

dx=⇒ dy

dx=

1

cos y.

We would like to express this in terms of x. Since x = sin y, wehave

cos y =√

1− x2 =⇒ d

dxarcsin x =

1√1− x2

.

Arcsin

Remember that


sin x

Example



1 = cos ydy

dx=⇒ dy

dx=

1

cos y.


cos y =√

1− x2 =⇒ d

dxarcsin x =

1√1− x2

.


Page10of13

Arcsin

Remember that


sin x

Example



1 = cos ydy

dx=⇒ dy

dx=

1

cos y.


cos y =√

1− x2 =⇒ d

dxarcsin x =

1√1− x2

.

Arccos

Example

Find the derivative of arccos.

Answer

d

dx(arccos x) =

−1√1− x2

Arccos

Example

Find the derivative of arccos.

Answer

d

dx(arccos x) =

−1√1− x2


Page12of13

Arctan

Example

Find the derivative of arctan.

SolutionFrom y = arctan x we have

x = tan y =⇒ 1 = sec2 ydy

dx,

sody

dx=

1

sec2 y.

But for all y

1 + tan2 y = sec2 y , =⇒ dy

dx=

1

1 + x2.

Arctan

Example




dx,

sody

dx=

1

sec2 y.

But for all y

1 + tan2 y = sec2 y , =⇒ dy

dx=

1

1 + x2.

Arctan

Example




dx,

sody

dx=

1

sec2 y.

But for all y

1 + tan2 y = sec2 y , =⇒ dy

dx=

1

1 + x2.

Arcsec

Example

Find the derivative of arcsec.

SolutionIf y = arcsec x then x = sec y, so

1 = sec y tan ydy

dx.

Thusdy

dx=

1

x tan y=

1

x√

x2 − 1.

Arcsec

Example

Find the derivative of arcsec.

SolutionIf y = arcsec x then x = sec y, so

1 = sec y tan ydy

dx.

Thusdy

dx=

1

x tan y=

1

x√

x2 − 1.

Business

Lesson 16: Implicit Differentiation