Implicit Differentiation and InverseTrigonometric Functions

MATH 161 Calculus I

J. Robert Buchanan

Department of Mathematics

Summer 2019

Explicit vs. Implicit Functions

-1.0 -0.5 0.0 0.5 1.0-4

-3

-2

-1

0

x

y

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

xy

y = x2 + 2x − 3 x2 + y2 = 1

Motivation (1 of 2)

Think of y as a function y(x) then

x2 + y2 = 1x2 + (y(x))2 = 1

(y(x))2 = 1− x2

y(x) =√

1− x2 or

y(x) = −√

1− x2.

The one equation defines two implicit functions of x .

Motivation (2 of 2)

ddx

[x2 + y2

]=

ddx

[1]

ddx

[x2 + (y(x))2

]= 0

2x + 2y(x)dydx

= 0

2ydydx

= −2x

dydx

= −xy

if y 6= 0.

Implicit Differentiation

The process of differentiating both sides of an equation isknown as implicit differentiation.

When we encounter a function of y , where y is implicitly afunction of x , we use the following derivative formula (the ChainRule):

ddx

[g(y)] = g′(y)dydx

Example

Finddydx

if x ey − 3y sin x = 1.

ddx

[x ey − 3y sin x ] =ddx

[1]

ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0

(xey − 3 sin x)y ′ = 3y cos x − ey

dydx

=3y cos x − ey

xey − 3 sin x

Example

Finddydx

if x ey − 3y sin x = 1.

ddx

[x ey − 3y sin x ] =ddx

[1]

ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0

(xey − 3 sin x)y ′ = 3y cos x − ey

dydx

=3y cos x − ey

xey − 3 sin x

Example

Finddydx

if x ey − 3y sin x = 1.

ddx

[x ey − 3y sin x ] =ddx

[1]

ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0

(xey − 3 sin x)y ′ = 3y cos x − ey

dydx

=3y cos x − ey

xey − 3 sin x

Example

Finddydx

if x ey − 3y sin x = 1.

ddx

[x ey − 3y sin x ] =ddx

[1]

ey + x eyy ′ − 3y ′ sin x − 3y cos x = 0(xey − 3 sin x)y ′ = 3y cos x − ey

dydx

=3y cos x − ey

xey − 3 sin x

ExampleFind the slope of the tangent line to the graph of

x3y2 − 4√

x = x2y at (x , y) = (4,1/2).

2 3 4 5 6

-1.0

-0.5

0.0

0.5

1.0

x

y

Solution

x3y2 − 4√

x = x2yddx

[x3y2 − 4x1/2

]=

ddx

[x2y

]

3x2y2 + 2x3y y ′ − 2√x

= 2xy + x2y ′

12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2

y ′ = − 748

Solution

x3y2 − 4√

x = x2yddx

[x3y2 − 4x1/2

]=

ddx

[x2y

]3x2y2 + 2x3y y ′ − 2√

x= 2xy + x2y ′

12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2

y ′ = − 748

Solution

x3y2 − 4√

x = x2yddx

[x3y2 − 4x1/2

]=

ddx

[x2y

]3x2y2 + 2x3y y ′ − 2√

x= 2xy + x2y ′

12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2

y ′ = − 748

Solution

x3y2 − 4√

x = x2yddx

[x3y2 − 4x1/2

]=

ddx

[x2y

]3x2y2 + 2x3y y ′ − 2√

x= 2xy + x2y ′

12 + 64y ′ − 1 = 4 + 16y ′ when x = 4, y = 1/2

y ′ = − 748

ExampleFind the equation of the tangent line to the graph of

x3

y+

y3

x= 1 at (x , y) = (1/2,0.746679).

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

x

y

Solution (1 of 2)

x3

y+

y3

x= 1

ddx

[x3

y+

y3

x

]=

ddx

[1]

3x2y − x3y ′

y2 +3y2y ′x − y3

x2 = 0

y ′ = 0.211708

when (x , y) = (1/2,0.746679).

Solution (1 of 2)

x3

y+

y3

x= 1

ddx

[x3

y+

y3

x

]=

ddx

[1]

3x2y − x3y ′

y2 +3y2y ′x − y3

x2 = 0

y ′ = 0.211708

when (x , y) = (1/2,0.746679).

Solution (1 of 2)

x3

y+

y3

x= 1

ddx

[x3

y+

y3

x

]=

ddx

[1]

3x2y − x3y ′

y2 +3y2y ′x − y3

x2 = 0

y ′ = 0.211708

when (x , y) = (1/2,0.746679).

Solution (2 of 2)

m =y − y0

x − x0

0.211708 =y − 0.746679

x − 1/2y = 0.211708x + 0.640825

Example

Find the horizontal and vertical tangents to the graph of theequation

xy2 − 2y = 2.

First we must find y ′:

ddx

[xy2 − 2y

]=

ddx

[2]

y2 + 2xy y ′ − 2y ′ = 0y2 + 2(xy − 1)y ′ = 0

y ′ =y2

2(1− xy)

Example

Find the horizontal and vertical tangents to the graph of theequation

xy2 − 2y = 2.

First we must find y ′:

ddx

[xy2 − 2y

]=

ddx

[2]

y2 + 2xy y ′ − 2y ′ = 0y2 + 2(xy − 1)y ′ = 0

y ′ =y2

2(1− xy)

Tangents

2 = xy2 − 2y

y ′ =y2

2(1− xy)

Tangent lines are horizontal when y ′ = 0 which implies y = 0.However, when y = 0 the first equation cannot be satisfied.Thus there are no points on the curve where the tangent line ishorizontal.

Tangent lines are vertical when y ′ is undefined. This impliesxy = 1 or equivalently x = 1/y . Substituting this into the firstequation yields

2 = y − 2y = −y =⇒ (x , y) = (−1/2,−2).

Tangents

2 = xy2 − 2y

y ′ =y2

2(1− xy)

Tangent lines are horizontal when y ′ = 0 which implies y = 0.However, when y = 0 the first equation cannot be satisfied.Thus there are no points on the curve where the tangent line ishorizontal.

Tangent lines are vertical when y ′ is undefined. This impliesxy = 1 or equivalently x = 1/y . Substituting this into the firstequation yields

2 = y − 2y = −y =⇒ (x , y) = (−1/2,−2).

Illustration

2 = xy2 − 2y

-1.0 -0.5 0.0 0.5 1.0

-3.0

-2.5

-2.0

-1.5

-1.0

x

y

Example

Find y ′′ given that x3 + y3 = 1.

First we must find y ′.

ddx

[x3 + y3

]=

ddx

[1]

3x2 + 3y2y ′ = 0

y ′ = −x2

y2

Then we must differentiate a second time.

Example

Find y ′′ given that x3 + y3 = 1.

First we must find y ′.

ddx

[x3 + y3

]=

ddx

[1]

3x2 + 3y2y ′ = 0

y ′ = −x2

y2

Then we must differentiate a second time.

Example

Find y ′′ given that x3 + y3 = 1.

First we must find y ′.

ddx

[x3 + y3

]=

ddx

[1]

3x2 + 3y2y ′ = 0

y ′ = −x2

y2

Then we must differentiate a second time.

Second DerivativeSo far we have:

x3 + y3 = 1

y ′ = −x2

y2

When we differentiate again we find,

ddx[y ′]

=ddx

[−x2

y2

]

y ′′ = −2xy2 − x2(2y)y ′

(y2)2 (quotient rule)

=−2xy2 + 2x2y(−x2/y2)

y4

=−2xy2 − 2x4/y

y4

=−2xy3 − 2x4

y5

Second DerivativeSo far we have:

x3 + y3 = 1

y ′ = −x2

y2

When we differentiate again we find,

ddx[y ′]

=ddx

[−x2

y2

]y ′′ = −2xy2 − x2(2y)y ′

(y2)2 (quotient rule)

=−2xy2 + 2x2y(−x2/y2)

y4

=−2xy2 − 2x4/y

y4

=−2xy3 − 2x4

y5

Second DerivativeSo far we have:

x3 + y3 = 1

y ′ = −x2

y2

When we differentiate again we find,

ddx[y ′]

=ddx

[−x2

y2

]y ′′ = −2xy2 − x2(2y)y ′

(y2)2 (quotient rule)

=−2xy2 + 2x2y(−x2/y2)

y4

=−2xy2 − 2x4/y

y4

=−2xy3 − 2x4

y5

Right Triangle Trigonometry

Suppose θ = sin−1 x or equivalently sin θ = x , then we canpicture θ as an acute angle in a right triangle.

1x

1- x2

θ=sin-1x

Derivative of f (x) = sin−1 x

We can use the Chain Rule to find derivatives of the inversetrigonometric functions.

sin(sin−1 x) = xddx

[sin(sin−1 x)

]=

ddx

[x ]

cos(sin−1 x)ddx

[sin−1 x

]= 1√

1− x2 ddx

[sin−1 x

]= 1

ddx

[sin−1 x

]=

1√1− x2

Derivatives of the Inverse Trigonometric Functions

ddx

[sin−1 x

]=

1√1− x2

ddx

[cos−1 x

]= − 1√

1− x2

ddx

[tan−1 x

]=

11 + x2

ddx

[sec−1 x

]=

1|x |√

x2 − 1ddx

[cot−1 x

]= − 1

1 + x2

ddx

[csc−1 x

]= − 1

|x |√

x2 − 1

Examples

Find the following derivatives:

1.ddx

[cos−1 3x

]

= − 3√1− 9x2

2.ddx

[sin−1 ex

]

=ex

√1− e2x

3.ddx

[tan−1 x2

]

=2x

1 + x4

4.ddx

[sec−1 ln x

]

=1/x

| ln x |√(ln x)2 − 1

Examples

Find the following derivatives:

1.ddx

[cos−1 3x

]= − 3√

1− 9x2

2.ddx

[sin−1 ex

]

=ex

√1− e2x

3.ddx

[tan−1 x2

]

=2x

1 + x4

4.ddx

[sec−1 ln x

]

=1/x

| ln x |√(ln x)2 − 1

Examples

Find the following derivatives:

1.ddx

[cos−1 3x

]= − 3√

1− 9x2

2.ddx

[sin−1 ex

]=

ex√

1− e2x

3.ddx

[tan−1 x2

]

=2x

1 + x4

4.ddx

[sec−1 ln x

]

=1/x

| ln x |√(ln x)2 − 1

Examples

Find the following derivatives:

1.ddx

[cos−1 3x

]= − 3√

1− 9x2

2.ddx

[sin−1 ex

]=

ex√

1− e2x

3.ddx

[tan−1 x2

]=

2x1 + x4

4.ddx

[sec−1 ln x

]

=1/x

| ln x |√(ln x)2 − 1

Examples

Find the following derivatives:

1.ddx

[cos−1 3x

]= − 3√

1− 9x2

2.ddx

[sin−1 ex

]=

ex√

1− e2x

3.ddx

[tan−1 x2

]=

2x1 + x4

4.ddx

[sec−1 ln x

]=

1/x| ln x |

√(ln x)2 − 1

Homework

I Read Section 2.8I Exercises: 1–27 odd (implicit differentiation), 29–37 odd

(inverse trigonometric functions)