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In this section, we will learn: How functions are defined implicitly.

1.
3

DIFFERENTIATION RULES

DIFFERENTIATION RULES

2.
DIFFERENTIATION RULES

The functions that we have met so far can be

described by expressing one variable

explicitly in terms of another variable.

For example, y = x 3 + 1 , or y = x sin x,

or in general y = f(x).

The functions that we have met so far can be

described by expressing one variable

explicitly in terms of another variable.

For example, y = x 3 + 1 , or y = x sin x,

or in general y = f(x).

3.
DIFFERENTIATION RULES

However, some functions are

defined implicitly.

However, some functions are

defined implicitly.

4.
DIFFERENTIATION RULES

3.5

Implicit Differentiation

In this section, we will learn:

How functions are defined implicitly.

3.5

Implicit Differentiation

In this section, we will learn:

How functions are defined implicitly.

5.
IMPLICIT DIFFERENTIATION Equations 1 and 2

Some examples of implicit functions

x2 + y2 = 25

x3 + y3 = 6xy

Some examples of implicit functions

x2 + y2 = 25

x3 + y3 = 6xy

6.
IMPLICIT DIFFERENTIATION

In some cases, it is possible to solve such an

equation for y as an explicit function (or

several functions) of x.

For instance, if we solve Equation 1 for y,

we get y = ± 25 −x 2

So, two of the functions determined by

the implicit Equation 1 are f ( x) = 25 −x 2

and g ( x) = − 25 −x 2

In some cases, it is possible to solve such an

equation for y as an explicit function (or

several functions) of x.

For instance, if we solve Equation 1 for y,

we get y = ± 25 −x 2

So, two of the functions determined by

the implicit Equation 1 are f ( x) = 25 −x 2

and g ( x) = − 25 −x 2

7.
IMPLICIT DIFFERENTIATION

The graphs of f and g are the upper

and lower semicircles of the circle

x2 + y2 = 25.

The graphs of f and g are the upper

and lower semicircles of the circle

x2 + y2 = 25.

8.
IMPLICIT DIFFERENTIATION

It’s not easy to solve Equation 2 for y

explicitly as a function of x by hand.

A computer algebra system has no trouble.

However, the expressions it obtains are

very complicated.

It’s not easy to solve Equation 2 for y

explicitly as a function of x by hand.

A computer algebra system has no trouble.

However, the expressions it obtains are

very complicated.

9.
FOLIUM OF DESCARTES

Nonetheless, Equation 2 is the equation

of a curve called the folium of Descartes

shown here and it implicitly defines y as

several functions of x.

Nonetheless, Equation 2 is the equation

of a curve called the folium of Descartes

shown here and it implicitly defines y as

several functions of x.

10.
FOLIUM OF DESCARTES

The graphs of three functions defined by

the folium of Descartes are shown.

The graphs of three functions defined by

the folium of Descartes are shown.

11.
IMPLICIT DIFFERENTIATION

When we say that f is a function defined

implicitly by Equation 2, we mean that

the equation x3 + [f(x)]3 = 6x f(x) is true for

all values of x in the domain of f.

When we say that f is a function defined

implicitly by Equation 2, we mean that

the equation x3 + [f(x)]3 = 6x f(x) is true for

all values of x in the domain of f.

12.
IMPLICIT DIFFERENTIATION

Fortunately, we don’t need to solve

an equation for y in terms of x to find

the derivative of y.

Fortunately, we don’t need to solve

an equation for y in terms of x to find

the derivative of y.

13.
IMPLICIT DIFFERENTIATION METHOD

Instead, we can use the method of

implicit differentiation.

This consists of differentiating both sides of

the equation with respect to x and then solving

the resulting equation for y’.

Instead, we can use the method of

implicit differentiation.

This consists of differentiating both sides of

the equation with respect to x and then solving

the resulting equation for y’.

14.
IMPLICIT DIFFERENTIATION METHOD

In the examples, it is always assumed that

the given equation determines y implicitly as

a differentiable function of x so that the

method of implicit differentiation can be

In the examples, it is always assumed that

the given equation determines y implicitly as

a differentiable function of x so that the

method of implicit differentiation can be

15.
IMPLICIT DIFFERENTIATION Example 1

dy

a. If x2 + y2 = 25, find .

dx

b. Find an equation of the tangent to

the circle x2 + y2 = 25 at the point (3, 4).

dy

a. If x2 + y2 = 25, find .

dx

b. Find an equation of the tangent to

the circle x2 + y2 = 25 at the point (3, 4).

16.
IMPLICIT DIFFERENTIATION Example 1 a

Differentiate both sides of the equation

x2 + y2 = 25:

d 2 2 d

( x + y ) = (25)

dx dx

d 2 d 2

(x ) + ( y ) = 0

dx dx

Differentiate both sides of the equation

x2 + y2 = 25:

d 2 2 d

( x + y ) = (25)

dx dx

d 2 d 2

(x ) + ( y ) = 0

dx dx

17.
IMPLICIT DIFFERENTIATION Example 1 a

Remembering that y is a function of x and

using the Chain Rule, we have:

d 2 d 2 dy dy

(y ) = (y ) = 2y

dx dy dx dx

dy

2x + 2 y =0

dx

dy dy x

Then, we solve this equation for : =−

dx dx y

Remembering that y is a function of x and

using the Chain Rule, we have:

d 2 d 2 dy dy

(y ) = (y ) = 2y

dx dy dx dx

dy

2x + 2 y =0

dx

dy dy x

Then, we solve this equation for : =−

dx dx y

18.
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 1

At the point (3, 4) we have x = 3 and y = 4.

So, dy = −3

dx 4

Thus, an equation of the tangent to the circle at (3, 4)

is: y – 4 = – ¾(x – 3) or 3x + 4y = 25.

At the point (3, 4) we have x = 3 and y = 4.

So, dy = −3

dx 4

Thus, an equation of the tangent to the circle at (3, 4)

is: y – 4 = – ¾(x – 3) or 3x + 4y = 25.

19.
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2

Solving the equation x2 + y2 = 25,

2

we get: y = ± 25 −x

The point (3, 4) lies on the upper semicircle y = 25 −x 2

So, we consider the function f ( x) = 25 −x 2

Solving the equation x2 + y2 = 25,

2

we get: y = ± 25 −x

The point (3, 4) lies on the upper semicircle y = 25 −x 2

So, we consider the function f ( x) = 25 −x 2

20.
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2

Differentiating f using the Chain Rule,

we have:

2 −1/ 2d 2

f '( x) = (25 −x )

1

2 (25 −x )

dx

2 −1/ 2

= 2 (25 −x ) (−2 x)

1

x

=−

2

25 −x

Differentiating f using the Chain Rule,

we have:

2 −1/ 2d 2

f '( x) = (25 −x )

1

2 (25 −x )

dx

2 −1/ 2

= 2 (25 −x ) (−2 x)

1

x

=−

2

25 −x

21.
IMPLICIT DIFFERENTIATION E. g. 1 b—Solution 2

So, 3

3

f '(3) = − =−

25 −32 4

As in Solution 1, an equation of the tangent is

3x + 4y = 25.

So, 3

3

f '(3) = − =−

25 −32 4

As in Solution 1, an equation of the tangent is

3x + 4y = 25.

22.
NOTE 1

The expression dy/dx = -x/y in Solution 1

gives the derivative in terms of both x and y.

It is correct no matter which function y is

determined by the given equation.

The expression dy/dx = -x/y in Solution 1

gives the derivative in terms of both x and y.

It is correct no matter which function y is

determined by the given equation.

23.
NOTE 1

2

For instance, for y = f ( x) = 25 −x ,

we have: dy x x

=− =−

dx y 25 −x 2

However, for y = g ( x) = − 25 −x 2 ,

we have:

dy x x x

=− =− =

dx y − 25 −x 2

25 −x 2

2

For instance, for y = f ( x) = 25 −x ,

we have: dy x x

=− =−

dx y 25 −x 2

However, for y = g ( x) = − 25 −x 2 ,

we have:

dy x x x

=− =− =

dx y − 25 −x 2

25 −x 2

24.
IMPLICIT DIFFERENTIATION Example 2

a. Find y’ if x3 + y3 = 6xy.

b. Find the tangent to the folium of Descartes

x3 + y3 = 6xy at the point (3, 3).

c. At what points in the first quadrant is

the tangent line horizontal?

a. Find y’ if x3 + y3 = 6xy.

b. Find the tangent to the folium of Descartes

x3 + y3 = 6xy at the point (3, 3).

c. At what points in the first quadrant is

the tangent line horizontal?

25.
IMPLICIT DIFFERENTIATION Example 2 a

Differentiating both sides of x3 + y3 = 6xy

with respect to x, regarding y as a function

of x, and using the Chain Rule on y3 and

the Product Rule on 6xy, we get:

3x2 + 3y2y’ = 6xy’ + 6y

or x2 + y2y’ = 2xy’ + 2y

Differentiating both sides of x3 + y3 = 6xy

with respect to x, regarding y as a function

of x, and using the Chain Rule on y3 and

the Product Rule on 6xy, we get:

3x2 + 3y2y’ = 6xy’ + 6y

or x2 + y2y’ = 2xy’ + 2y

26.
IMPLICIT DIFFERENTIATION Example 2 a

Now, we solve for y’:

2 2

y y '−2 xy ' = 2 y −x

2 2

( y −2 x) y ' = 2 y −x

2

2 y −x

y' = 2

y −2 x

Now, we solve for y’:

2 2

y y '−2 xy ' = 2 y −x

2 2

( y −2 x) y ' = 2 y −x

2

2 y −x

y' = 2

y −2 x

27.
IMPLICIT DIFFERENTIATION Example 2 b

2

When x = y = 3, 2 ⋅3 −3

y'= 2 = −1

3 −2 ⋅3

A glance at the figure confirms

that this is a reasonable value

for the slope at (3, 3).

So, an equation of the tangent

to the folium at (3, 3) is:

y – 3 = – 1(x – 3) or x + y = 6.

2

When x = y = 3, 2 ⋅3 −3

y'= 2 = −1

3 −2 ⋅3

A glance at the figure confirms

that this is a reasonable value

for the slope at (3, 3).

So, an equation of the tangent

to the folium at (3, 3) is:

y – 3 = – 1(x – 3) or x + y = 6.

28.
IMPLICIT DIFFERENTIATION Example 2 c

The tangent line is horizontal if y’ = 0.

Using the expression for y’ from (a), we see that y’ = 0

when 2y – x2 = 0 (provided that y2 – 2x ≠ 0).

Substituting y = ½x2 in the equation of the curve,

we get x3 + (½x2)3 = 6x(½x2) which simplifies to

x6 = 16x3.

The tangent line is horizontal if y’ = 0.

Using the expression for y’ from (a), we see that y’ = 0

when 2y – x2 = 0 (provided that y2 – 2x ≠ 0).

Substituting y = ½x2 in the equation of the curve,

we get x3 + (½x2)3 = 6x(½x2) which simplifies to

x6 = 16x3.

29.
IMPLICIT DIFFERENTIATION Example 2 c

Since x ≠ 0 in the first quadrant,

we have x3 = 16.

If x = 161/3 = 24/3, then y = ½(28/3) = 25/3.

Since x ≠ 0 in the first quadrant,

we have x3 = 16.

If x = 161/3 = 24/3, then y = ½(28/3) = 25/3.

30.
IMPLICIT DIFFERENTIATION Example 2 c

Thus, the tangent is horizontal at (0, 0)

and at (24/3, 25/3), which is approximately

(2.5198, 3.1748).

Looking at the figure, we see

that our answer is reasonable.

Thus, the tangent is horizontal at (0, 0)

and at (24/3, 25/3), which is approximately

(2.5198, 3.1748).

Looking at the figure, we see

that our answer is reasonable.

31.
NOTE 2

There is a formula for the three roots

of a cubic equation that is like

the quadratic formula, but much more

There is a formula for the three roots

of a cubic equation that is like

the quadratic formula, but much more

32.
NOTE 2

If we use this formula (or a computer algebra

system) to solve the equation x3 + y3 = 6xy

for y in terms of x, we get three functions

determined by the following equation.

If we use this formula (or a computer algebra

system) to solve the equation x3 + y3 = 6xy

for y in terms of x, we get three functions

determined by the following equation.

33.
NOTE 2

y = f ( x) = 3 −12 x 3 + 1

4 x 6 −8 x 3 + 3 −12 x 3 − 1

4 x 6 −8 x3

⎡ ⎤

y = ⎢−f ( x) ± −3 ⎛

1

2

3

−1 3

⎜ 2x +

1

4

6 3

x −8 x − − x −

3 1

2

3 1

4

6 3⎞

x −8 x ⎟⎥

⎣ ⎝ ⎠⎦

y = f ( x) = 3 −12 x 3 + 1

4 x 6 −8 x 3 + 3 −12 x 3 − 1

4 x 6 −8 x3

⎡ ⎤

y = ⎢−f ( x) ± −3 ⎛

1

2

3

−1 3

⎜ 2x +

1

4

6 3

x −8 x − − x −

3 1

2

3 1

4

6 3⎞

x −8 x ⎟⎥

⎣ ⎝ ⎠⎦

34.
NOTE 2

These are the three functions whose

graphs are shown in the earlier figure.

These are the three functions whose

graphs are shown in the earlier figure.

35.
NOTE 2

You can see that the method of implicit

differentiation saves an enormous amount of

work in cases such as this.

You can see that the method of implicit

differentiation saves an enormous amount of

work in cases such as this.

36.
NOTE 2

Moreover, implicit differentiation works just

as easily for equations such as

y5 + 3x2y2 + 5x4 = 12

for which it is impossible to find a similar

expression for y in terms of x.

Moreover, implicit differentiation works just

as easily for equations such as

y5 + 3x2y2 + 5x4 = 12

for which it is impossible to find a similar

expression for y in terms of x.

37.
IMPLICIT DIFFERENTIATION Example 3

Find y’ if sin(x + y) = y2 cos x.

Differentiating implicitly with respect to x and

remembering that y is a function of x, we get:

cos( x + y ) ⋅(1 + y ') = y 2 (−sin x) + (cos x)(2 yy ')

Note that we have used the Chain Rule on the left side

and the Product Rule and Chain Rule on the right side.

Find y’ if sin(x + y) = y2 cos x.

Differentiating implicitly with respect to x and

remembering that y is a function of x, we get:

cos( x + y ) ⋅(1 + y ') = y 2 (−sin x) + (cos x)(2 yy ')

Note that we have used the Chain Rule on the left side

and the Product Rule and Chain Rule on the right side.

38.
IMPLICIT DIFFERENTIATION Example 3

If we collect the terms that involve y’,

we get:

cos( x + y ) + y 2 sin x = (2 y cos x) y '−cos( x + y ) ⋅y '

y 2 sin x + cos( x + y )

So, y ' =

2 y cos x −cos( x + y )

If we collect the terms that involve y’,

we get:

cos( x + y ) + y 2 sin x = (2 y cos x) y '−cos( x + y ) ⋅y '

y 2 sin x + cos( x + y )

So, y ' =

2 y cos x −cos( x + y )

39.
IMPLICIT DIFFERENTIATION Example 3

The figure, drawn with the implicit-plotting

command of a computer algebra system,

shows part of the curve sin(x + y) = y2 cos x.

As a check on our calculation,

notice that y’ = -1 when

x = y = 0 and it appears that

the slope is approximately -1

at the origin.

The figure, drawn with the implicit-plotting

command of a computer algebra system,

shows part of the curve sin(x + y) = y2 cos x.

As a check on our calculation,

notice that y’ = -1 when

x = y = 0 and it appears that

the slope is approximately -1

at the origin.

40.
IMPLICIT DIFFERENTIATION

The following example shows how to

find the second derivative of a function

that is defined implicitly.

The following example shows how to

find the second derivative of a function

that is defined implicitly.

41.
IMPLICIT DIFFERENTIATION Example 4

Find y” if x4 + y4 = 16.

Differentiating the equation implicitly with

respect to x, we get 4x3 + 4y3y’ = 0.

Find y” if x4 + y4 = 16.

Differentiating the equation implicitly with

respect to x, we get 4x3 + 4y3y’ = 0.

42.
IMPLICIT DIFFERENTIATION E. g. 4—Equation 3

Solving for y’ gives:

3

x

y' = − 3

y

Solving for y’ gives:

3

x

y' = − 3

y

43.
IMPLICIT DIFFERENTIATION Example 4

To find y’’, we differentiate this expression

for y’ using the Quotient Rule and

remembering that y is a function of x:

d ⎛ x3 ⎞ y 3 (d / dx)( x 3 ) −x 3 (d / dx)( y 3 )

y '' = ⎜− 3 ⎟ = − 3 2

dx ⎝ y ⎠ (y )

y 3 ⋅3 x 2 −x 3 (3 y 2 y ')

=− 6

y

To find y’’, we differentiate this expression

for y’ using the Quotient Rule and

remembering that y is a function of x:

d ⎛ x3 ⎞ y 3 (d / dx)( x 3 ) −x 3 (d / dx)( y 3 )

y '' = ⎜− 3 ⎟ = − 3 2

dx ⎝ y ⎠ (y )

y 3 ⋅3 x 2 −x 3 (3 y 2 y ')

=− 6

y

44.
IMPLICIT DIFFERENTIATION Example 4

If we now substitute Equation 3 into

this expression, we get:

2 3 3 2 ⎛ x ⎞

3

3 x y −3 x y ⎜− 3 ⎟

⎝ y ⎠

y '' = − 6

y

2 4 6 2 4 4

3( x y + x ) 3x ( y + x )

=− 7

=− 7

y y

If we now substitute Equation 3 into

this expression, we get:

2 3 3 2 ⎛ x ⎞

3

3 x y −3 x y ⎜− 3 ⎟

⎝ y ⎠

y '' = − 6

y

2 4 6 2 4 4

3( x y + x ) 3x ( y + x )

=− 7

=− 7

y y

45.
IMPLICIT DIFFERENTIATION Example 4

However, the values of x and y must satisfy

the original equation x4 + y4 = 16.

So, the answer simplifies to:

2 2

3 x (16) x

y '' = − 7

= −48 7

y y

However, the values of x and y must satisfy

the original equation x4 + y4 = 16.

So, the answer simplifies to:

2 2

3 x (16) x

y '' = − 7

= −48 7

y y

46.
INVERSE TRIGONOMETRIC FUNCTIONS (ITFs)

The inverse trigonometric functions were

reviewed in Section 1.6

We discussed their continuity in Section 2.5 and

their asymptotes in Section 2.6

The inverse trigonometric functions were

reviewed in Section 1.6

We discussed their continuity in Section 2.5 and

their asymptotes in Section 2.6

47.
DERIVATIVES OF ITFs

Here, we use implicit differentiation to find

the derivatives of the inverse trigonometric

functions—assuming that these functions are

Here, we use implicit differentiation to find

the derivatives of the inverse trigonometric

functions—assuming that these functions are

48.
DERIVATIVES OF ITFs

In fact, if f is any one-to-one differentiable

function, it can be proved that its inverse

function f -1 is also differentiable—except

where its tangents are vertical.

This is plausible because the graph of a differentiable

function has no corner or kink.

So, if we reflect it about y = x, the graph of its inverse

function also has no corner or kink.

In fact, if f is any one-to-one differentiable

function, it can be proved that its inverse

function f -1 is also differentiable—except

where its tangents are vertical.

This is plausible because the graph of a differentiable

function has no corner or kink.

So, if we reflect it about y = x, the graph of its inverse

function also has no corner or kink.

49.
DERIVATIVE OF ARCSINE FUNCTION

Recall the definition of the arcsine function:

−1 π π

y = sin x means sin y = x and − ≤ y ≤

2 2

Differentiating sin y = x implicitly with respect

to x, we obtain:

dy dy 1

cos y = 1 or =

dx dx cos y

Recall the definition of the arcsine function:

−1 π π

y = sin x means sin y = x and − ≤ y ≤

2 2

Differentiating sin y = x implicitly with respect

to x, we obtain:

dy dy 1

cos y = 1 or =

dx dx cos y

50.
DERIVATIVE OF ARCSINE FUNCTION

Now, cos y ≥ 0, since –π/2 ≤ y ≤ π/2.

So, cos y = 1 −sin 2 y = 1 −x 2

Thus, dy = 1 = 1

dx cos y 1 −x 2

d −1 1

(sin x) =

dx 1 −x 2

Now, cos y ≥ 0, since –π/2 ≤ y ≤ π/2.

So, cos y = 1 −sin 2 y = 1 −x 2

Thus, dy = 1 = 1

dx cos y 1 −x 2

d −1 1

(sin x) =

dx 1 −x 2

51.
DERIVATIVE OF ARCTANGENT FUNCTION

The formula for the derivative of the

arctangent function is derived in a similar way.

If y = tan -1x, then tan y = x.

Differentiating this latter equation implicitly

with respect to x, we have:

2 dy

sec y =1

dx

dy 1 1 1

= = =

dx sec y 1 + tan y 1 + x 2

2 2

d −1 1

(tan x) =

dx 1 + x2

The formula for the derivative of the

arctangent function is derived in a similar way.

If y = tan -1x, then tan y = x.

Differentiating this latter equation implicitly

with respect to x, we have:

2 dy

sec y =1

dx

dy 1 1 1

= = =

dx sec y 1 + tan y 1 + x 2

2 2

d −1 1

(tan x) =

dx 1 + x2

52.
DERIVATIVES OF ITFs Example 5

1

a. y = −1

sin x

b. f(x) = x arctan x

1

a. y = −1

sin x

b. f(x) = x arctan x

53.
DERIVATIVES OF ITFs Example 5 a

dy d −1 −1 −1 −2 d −1

= (sin x) = −(sin x) (sin x)

dx dx dx

1

=−

−1 2 2

(sin x) 1 −x

dy d −1 −1 −1 −2 d −1

= (sin x) = −(sin x) (sin x)

dx dx dx

1

=−

−1 2 2

(sin x) 1 −x

54.
DERIVATIVES OF ITFs Example 5 b

1 −1/ 2

f '( x) = x 2

(

1

2 x ) + arctan x

1+ ( x )

x

= + arctan x

2(1 + x)

1 −1/ 2

f '( x) = x 2

(

1

2 x ) + arctan x

1+ ( x )

x

= + arctan x

2(1 + x)

55.
DERIVATIVES OF ITFs

The inverse trigonometric functions

that occur most frequently are the ones

that we have just discussed.

The inverse trigonometric functions

that occur most frequently are the ones

that we have just discussed.

56.
DERIVATIVES OF ITFs

The derivatives of the remaining four are

given in this table.

The proofs of the formulas are left as exercises.

d −1 1 d −1 1

(sin x) = (csc x) = −

dx 1 −x 2 dx x x 2 −1

d −1 1 d −1 1

(cos x) = − (sec x) =

dx 1 −x 2 dx x x 2 −1

d −1 1 d −1 1

(tan x) = (cot x) = −

dx 1 + x2 dx 1 + x2

The derivatives of the remaining four are

given in this table.

The proofs of the formulas are left as exercises.

d −1 1 d −1 1

(sin x) = (csc x) = −

dx 1 −x 2 dx x x 2 −1

d −1 1 d −1 1

(cos x) = − (sec x) =

dx 1 −x 2 dx x x 2 −1

d −1 1 d −1 1

(tan x) = (cot x) = −

dx 1 + x2 dx 1 + x2