Math 112 Lecture 13: Differentiation -- Derivatives of ...€¦ · Derivatives of the Inverse Trigonometric Functions Derivative of sin Derivative of cos Using the Chain Rule Derivative

Derivatives of the Basic Trigonometric FunctionsApplying the Trig Function Derivative Rules

Derivatives of the Inverse Trigonometric Functions

Lecture 13: Differentiation – Derivatives of Trigonometric Functions

Derivatives of the Basic Trigonometric FunctionsDerivative of sinDerivative of cos Using the Chain RuleDerivative of tan Using the Quotient RuleDerivatives the Six Trigonometric Functions

Applying the Trig Function Derivative RulesExample 46 – Differentiating with Trig FunctionsExample 47 – Damped Oscillations

Derivatives of the Inverse Trigonometric FunctionsThe arctan FunctionThe arcsin FunctionExample 48 – Differentiating with Inverse Trig Functions

Clint LeeMath 112 Lecture 13: Differentiation – Derivatives of Trigonometric Functions1/25



Derivative of sinDerivative of cos Using the Chain RuleDerivative of tan Using the Quotient RuleDerivatives the Six Trigonometric Functions

Derivative of sin

Recall that in Example 31(c) we guessed that

ddx

sin x = cos x

by considering the graphs of sin and cos.





Derivative of sin


ddx

sin x = cos x

by considering the graphs of sin and cos. We will now prove this using thedefinition of the derivative and some basic trigonometric identities.





Derivative of sin


ddx

sin x = cos x


First recall the sum and difference formulas for sin





Derivative of sin


ddx

sin x = cos x



sin (x ± y)





Derivative of sin


ddx

sin x = cos x



sin (x ± y) = sin x cos y ± cos x sin y





Derivative of sin


ddx

sin x = cos x




Though we don’t need it right away, the corresponding formula for cos is





Derivative of sin


ddx

sin x = cos x





cos (x ± y)





Derivative of sin


ddx

sin x = cos x





cos (x ± y) = cos x cos y ∓ sin x sin y





Derivative of sin – continued

With f (x) = sin x, using Formula 3 we have







f ′(x)







f ′(x) = limh→0

sin (x + h) − sin xh







f ′(x) = limh→0


= limh→0

sin x cos h + cos x sin h − sin xh







f ′(x) = limh→0


= limh→0


= limh→0

sin x (cos h − 1) + cos x sin hh







f ′(x) = limh→0


= limh→0


= limh→0


= sin x(

limh→0

cos h − 1h

)

+ cos x(

limh→0

sin hh

)







f ′(x) = limh→0


= limh→0


= limh→0


= sin x(

limh→0

cos h − 1h

)

+ cos x(

limh→0

sin hh

)

Recall that using the Squeeze Theorem we proved that

limx→0

sin xx







f ′(x) = limh→0


= limh→0


= limh→0


= sin x(

limh→0

cos h − 1h

)

+ cos x(

limh→0

sin hh

)

Recall that using the Squeeze Theorem we proved that

limx→0

sin xx

= 1






Further, using the same approach as used in Example 13 we can show that

limh→0

cos h − 1h







limh→0

cos h − 1h

= 0







limh→0

cos h − 1h

= 0

Thus we have

f ′(x) =ddx

sin x







limh→0

cos h − 1h

= 0

Thus we have

f ′(x) =ddx

sin x = sin x (0) + cos x (1)







limh→0

cos h − 1h

= 0

Thus we have

f ′(x) =ddx

sin x = sin x (0) + cos x (1) = cos x







limh→0

cos h − 1h

= 0

Thus we have

f ′(x) =ddx


as we predicted.







limh→0

cos h − 1h

= 0

Thus we have

f ′(x) =ddx


as we predicted.

You use the sum formula for cos to prove the corresponding differentiationformula for cos x, which is

ddx

cos x







limh→0

cos h − 1h

= 0

Thus we have

f ′(x) =ddx


as we predicted.

You use the sum formula for cos to prove the corresponding differentiationformula for cos x, which is

ddx

cos x = − sin x





Derivative of cos Using the Chain Rule

We can prove the formula for the derivative of cos in a different way. Twobasic trigonometric identities are







sin(

π

2 − x)







sin(

π

2 − x)

= cos x







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

=







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x

since x and π

2 − x are complementary angles in a right triangle.







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x

since x and π

2 − x are complementary angles in a right triangle.Thus, using the Chain Rule gives

ddx

cos x







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x

since x and π


ddx

cos x =d

dxsin(

π

2 − x)







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x

since x and π


ddx

cos x =d

dxsin(

π

2 − x)

= cos(

π

2 − x) d

dx(

π

2 − x)







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x

since x and π


ddx

cos x =d

dxsin(

π

2 − x)

= cos(

π

2 − x) d

dx(

π

2 − x)

= sin x(−1)







sin(

π

2 − x)

= cos xcos

(

π

2 − x)

= sin x

since x and π


ddx

cos x =d

dxsin(

π

2 − x)

= cos(

π

2 − x) d

dx(

π

2 − x)

= sin x(−1) = − sin x





Derivative of tan Using the Quotient Rule

Recall thattan x






Recall thattan x =

sin xcos x






Recall thattan x =

sin xcos x

Thus, using the Quotient Rule gives

ddx

tan x






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x

=

(

ddx

sin x)

cos x − sin x(

ddx

cos x)

(cos x)2






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x

=

(

ddx

sin x)

cos x − sin x(

ddx

cos x)

(cos x)2

=(cos x) cos x − sin x (− sin x)

cos2 x






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x

=

(

ddx

sin x)

cos x − sin x(

ddx

cos x)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x

=

(

ddx

sin x)

cos x − sin x(

ddx

cos x)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x=

1cos2 x






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x

=

(

ddx

sin x)

cos x − sin x(

ddx

cos x)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x=

1cos2 x

=

(

1cos x

)2






Recall thattan x =

sin xcos x


ddx

tan x =ddx

sin xcos x

=

(

ddx

sin x)

cos x − sin x(

ddx

cos x)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x=

1cos2 x

=

(

1cos x

)2= sec2 x





Derivative of tan Using the Quotient Rule – continued

An alternative way to simplify the previous expression is

ddx

tan x







ddx

tan x =cos2 x + sin2 x

cos2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x

So thatd

dxtan x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x

So thatd

dxtan x = sec2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x

So thatd

dxtan x = sec2 x = 1 + tan2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x

So thatd


The equality above can also be proved using the Pythagorean identity

1 + tan2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x

So thatd



1 + tan2 x = sec2 x







ddx


cos2 x=

cos2 xcos2 x

+sin2 xcos2 x

= 1 +

(

sin xcos x

)2= 1 + tan2 x

So thatd



1 + tan2 x = sec2 x

Most text books use the sec2 x formula for the derivative of tan x, but Mapleand other symbolic differentiating programs use the 1 + tan2 x formula.





Derivatives the Six Trigonometric Functions

Using basic differentiation rules as in the derivation of the derivative formulafor tan we can find derivative formulas for all of the other trigonometricfunctions.






Using basic differentiation rules as in the derivation of the derivative formulafor tan we can find derivative formulas for all of the other trigonometricfunctions. Also, recall that when we derived the General Rule for theExponential Function we stated that we would give all derivative formulas ina general form using the Chain Rule. In this form we introduce anintermediate variable u assumed to represent some function of x. With thisassumption the derivative rules for all six basic trigonometric functions are:







ddx

sin u







ddx

sin u = cos ududx







ddx

sin u = cos ududx

ddx

cos u







ddx

sin u = cos ududx

ddx

cos u = − sin ududx







ddx

sin u = cos ududx

ddx


ddx

tan u







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dx







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dxddx

sec u







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dxddx

sec u = sec u tan ududx







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dxddx


ddx

csc u







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dxddx


ddx

csc u = − csc u cot ududx







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dxddx


ddx


ddx

cot u







ddx

sin u = cos ududx

ddx


ddx

tan u = sec2 ududx

=(

1 + tan2 u) du

dxddx


ddx


ddx

cot u = − csc2 ududx

= −(

1 + cot2 u) du

dx




Example 46 – Differentiating with Trig FunctionsExample 47 – Damped Oscillations

Example 46 – Differentiating with Trig Functions

Find and simplify the indicated derivative(s) of each function.

(a) Find f ′(x) and f ′′(x) for f (x) = x2 cos (3x).

(b) Finddsdt

for s =cos t

sin t + cos t.

(c) Find C′(x) for C(x) = tan(

e√

1+x2)

.





Solution: Example 46(a)

Using the Product Rule followed by the Chain Rule (for cos (3x)) gives

f ′(x)







f ′(x) = 2x cos (3x) + x2 [−3 sin (3x)]







f ′(x) = 2x cos (3x) + x2 [−3 sin (3x)]

= 2x cos (3x) − 3x2 sin (3x)







f ′(x) = 2x cos (3x) + x2 [−3 sin (3x)]

= 2x cos (3x) − 3x2 sin (3x)

= x [2 cos (3x) − 3x sin (3x)]







f ′(x) = 2x cos (3x) + x2 [−3 sin (3x)]

= 2x cos (3x) − 3x2 sin (3x)

= x [2 cos (3x) − 3x sin (3x)]

For f ′′(x) use the expression in the second line. Again using the Product andChain Rules gives

f ′′(x)







f ′(x) = 2x cos (3x) + x2 [−3 sin (3x)]

= 2x cos (3x) − 3x2 sin (3x)

= x [2 cos (3x) − 3x sin (3x)]


f ′′(x) = 2 cos (3x) − 6x sin (3x) − 6x sin (3x) − 9x2 cos (3x)







f ′(x) = 2x cos (3x) + x2 [−3 sin (3x)]

= 2x cos (3x) − 3x2 sin (3x)

= x [2 cos (3x) − 3x sin (3x)]


f ′′(x) = 2 cos (3x) − 6x sin (3x) − 6x sin (3x) − 9x2 cos (3x)

=(

2 − 9x2)

cos (3x) − 12x sin (3x)





Solution: Example 46(b)

Using the Quotient Rule gives

dsdt







dsdt

=− sin t (sin t + cos t) − cos t (cos t − sin t)

(sin t + cos t)2







dsdt


(sin t + cos t)2

=− sin2 t − sin t cos t − cos2 t + cos t sin t

(sin t + cos t)2







dsdt


(sin t + cos t)2


(sin t + cos t)2

=−(

sin2 t + cos2 t)

(sin t + cos t)2







dsdt


(sin t + cos t)2


(sin t + cos t)2

=−(

sin2 t + cos2 t)

(sin t + cos t)2

= − 1

(sin t + cos t)2







dsdt


(sin t + cos t)2


(sin t + cos t)2

=−(

sin2 t + cos2 t)

(sin t + cos t)2

= − 1

(sin t + cos t)2

This example illustrates the fact that when simplifying derivatives involvingtrig functions, you sometimes need to use standard trigonometric identities.





Solution: Example 46(c)

This is a composite function with







C(x)







C(x) = f (g (h (k(x))))







C(x) = f (g (h (k(x))))

where

f (x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x,







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex,







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2,







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2, k(x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2, k(x) = 1 + x2







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2, k(x) = 1 + x2

Using the Chain Rule three times gives

C′(x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2, k(x) = 1 + x2


C′(x) = f ′ (g (h (k(x)))) g′ (h (k(x))) h′ (k(x)) k′(x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2, k(x) = 1 + x2



= sec2(

e√

1+x2) (

e√

1+x2) (

12

) (

1 + x2)−1/2

(2x)







C(x) = f (g (h (k(x))))

where

f (x) = tan x, g(x) = ex, h(x) =√

x = x1/2, k(x) = 1 + x2



= sec2(

e√

1+x2) (

e√

1+x2) (

12

) (

1 + x2)−1/2

(2x)

=xe

√1+x2 sec2

(

e√

1+x2)

√1 + x2





Example 47 – Damped Oscillations

Consider the functionq(t) = e−7t sin (24t)

This function describes damped simple harmonic motion. It gives theposition of a mass attached to a spring relative to the equilibrium (resting)position of the spring. A frictional force acts to gradually slow the mass.

(a) Find q′(t) and q′′(t) and explain their meaning in terms of the dampedoscillatory motion.

(b) Note that q(0) = 0. This means that the initial position of the mass is atthe equilibrium position of the spring. Find the initial velocity of themass. Also find the velocity when the mass first returns to theequilibrium position.

(c) Draw a graph of the function q(t).(d) Find the first two times when the oscillating mass turns around. Show

the corresponding points on the graph of q(t).(e) Show that the function q(t) satisfies the differential equation

d2qdt2 + 14

dqdt

+ 625q = 0






Using the Product and Chain Rules gives







q′(t)







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]

From our interpretation of the derivative as a rate of change, we know thatthis is the velocity of the mass at time t.







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]

From our interpretation of the derivative as a rate of change, we know thatthis is the velocity of the mass at time t.Taking the derivative of the expression above for q′(t) gives







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]


q′′(t)







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]


q′′(t) = −7e−7t [24 cos (24t) − 7 sin (24t)] + e−7t[

−242 sin (24t) − 7(24) cos (24t)]







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]


q′′(t) = −7e−7t [24 cos (24t) − 7 sin (24t)] + e−7t[

−242 sin (24t) − 7(24) cos (24t)]

= −e−7t[

14(24) cos (24t) +(

242 − 72)

sin (24t)]







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]


q′′(t) = −7e−7t [24 cos (24t) − 7 sin (24t)] + e−7t[

−242 sin (24t) − 7(24) cos (24t)]

= −e−7t[

14(24) cos (24t) +(

242 − 72)

sin (24t)]

= −e−7t [336 cos (24t) + 527 sin (24t)]







q′(t) = −7e−7t sin (24t) + e−7t [24 cos (24t)]

= e−7t [24 cos (24t) − 7 sin (24t)]


q′′(t) = −7e−7t [24 cos (24t) − 7 sin (24t)] + e−7t[

−242 sin (24t) − 7(24) cos (24t)]

= −e−7t[

14(24) cos (24t) +(

242 − 72)

sin (24t)]

= −e−7t [336 cos (24t) + 527 sin (24t)]

The rate of change of velocity is acceleration, so this gives the acceleration ofthe mass at time t.






Substituting t = 0 into the expression for the velocity v(t) = q′(t) gives

v(0)







v(0) = e0 [24 cos (0) − 7 sin (0)]







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24

The mass returns to the equilibrium position when .







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24

The mass returns to the equilibrium position when q(t) = 0.







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24

The mass returns to the equilibrium position when q(t) = 0. The first timeafter t = 0 when this happens is when







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t =π

24







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t =π

24

The velocity at this time is

v(

π

24

)







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t =π

24


v(

π

24

)

= e−7π/24 [24 cos (π) − 7 sin (π)]







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t =π

24


v(

π

24

)

= e−7π/24 [24 cos (π) − 7 sin (π)] = −24e−7π/24







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t =π

24


v(

π

24

)

= e−7π/24 [24 cos (π) − 7 sin (π)] = −24e−7π/24 = −9.6







v(0) = e0 [24 cos (0) − 7 sin (0)] = 24


24t = π ⇒ t =π

24


v(

π

24

)

= e−7π/24 [24 cos (π) − 7 sin (π)] = −24e−7π/24 = −9.6

This velocity is less than the initial velocity and in the opposite direction.






The graph of the function q(t) looks like this.The amplitude of the motion decreasesfollowing an envelope given by the decayingexponential function e−7t, as shown.

0.1 0.2 0.3 0.4






The graph of the function q(t) looks like this.The amplitude of the motion decreasesfollowing an envelope given by the decayingexponential function e−7t, as shown. As wesaw in the last part, not only does theamplitude decrease, but so does the velocity.

0.1 0.2 0.3 0.4






The graph of the function q(t) looks like this.The amplitude of the motion decreasesfollowing an envelope given by the decayingexponential function e−7t, as shown. As wesaw in the last part, not only does theamplitude decrease, but so does the velocity.Further, note that where the graph is concavedown, q′′(t) < 0, the mass is decelerating. Thevelocity is getting less positive or morenegative.

0.1 0.2 0.3 0.4






The graph of the function q(t) looks like this.The amplitude of the motion decreasesfollowing an envelope given by the decayingexponential function e−7t, as shown. As wesaw in the last part, not only does theamplitude decrease, but so does the velocity.Further, note that where the graph is concavedown, q′′(t) < 0, the mass is decelerating. Thevelocity is getting less positive or morenegative. And, where the graph is concaveup, q′′(t) > 0, the mass is accelerating. Thevelocity is getting more positive or lessnegative.

0.1 0.2 0.3 0.4





Solution: Example 47(d)

The mass turns around when v(t) = q′(t) = 0.This happens when







24 cos (24t) − 7 sin (24t) = 0

0.1 0.2 0.3 0.4







24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =247

0.1 0.2 0.3 0.4







24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =247

⇒24t = arctan(

247

)

+ kπ

0.1 0.2 0.3 0.4







24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =247

⇒24t = arctan(

247

)

+ kπ

where k is any integer.

0.1 0.2 0.3 0.4







24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =247

⇒24t = arctan(

247

)

+ kπ

where k is any integer. The first two positive tvalues are t1 = 0.0536 and t2 = 0.1845.

0.1 0.2 0.3 0.4







24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =247

⇒24t = arctan(

247

)

+ kπ

where k is any integer. The first two positive tvalues are t1 = 0.0536 and t2 = 0.1845. Atthese values the tangent line to the graph ofq(t) is horizontal, as shown.

0.1 0.2 0.3 0.4







24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =247

⇒24t = arctan(

247

)

+ kπ

where k is any integer. The first two positive tvalues are t1 = 0.0536 and t2 = 0.1845. Atthese values the tangent line to the graph ofq(t) is horizontal, as shown.

0.1 0.2 0.3 0.4





Solution: Example 47(e)

From part (a) we have

d2qdt2 + 14

dqdt

+ 625q







d2qdt2 + 14

dqdt

+ 625q

= −e−7t [336 cos (24t) + 527 sin (24t)]







d2qdt2 + 14

dqdt

+ 625q

= −e−7t [336 cos (24t) + 527 sin (24t)]

+ 14e−7t [24 cos (24t) − 7 sin (24t)] + 625e−7t sin (24t)







d2qdt2 + 14

dqdt

+ 625q

= −e−7t [336 cos (24t) + 527 sin (24t)]


= e−7t [(−336 + 14 × 24) cos (24t) + (−527 − 14 × 7 + 625) sin (24t)]







d2qdt2 + 14

dqdt

+ 625q

= −e−7t [336 cos (24t) + 527 sin (24t)]


= e−7t [(−336 + 14 × 24) cos (24t) + (−527 − 14 × 7 + 625) sin (24t)]= 0




The arctan FunctionThe arcsin FunctionExample 48 – Differentiating with Inverse Trig Functions

The arctan Function

Recall that the graph of the tangent functionlooks like this.





The arctan Function

Recall that the graph of the tangent functionlooks like this. From this graph we realize thatthe tangent function is not one-to-one

π

2−π

2





The arctan Function

Recall that the graph of the tangent functionlooks like this. From this graph we realize thatthe tangent function is not one-to-one, and sodoes not have an inverse function.

π

2−π

2





The arctan Function

Recall that the graph of the tangent functionlooks like this. From this graph we realize thatthe tangent function is not one-to-one, and sodoes not have an inverse function. However,restricting the domain of the tangent function, π

2−π

2





The arctan Function

Recall that the graph of the tangent functionlooks like this. From this graph we realize thatthe tangent function is not one-to-one, and sodoes not have an inverse function. However,restricting the domain of the tangent function,as shown in the graph, to the interval

− π

2< x <

π

2

π

2−π

2





The arctan Function


− π

2< x <

π

2

gives a one-to-one function.

π

2−π

2





The arctan Function


− π

2< x <

π

2


π

2−π

2

So the inverse of the tangent function is defined as

arctan x = tan−1 x = the angle between − π

2 and π

2 whose tangent is x





The Derivative of the arctan Function

With this definition of the inverse tangent function, we see that its domain is






With this definition of the inverse tangent function, we see that its domain isall real numbers and its range is







(

−π

2 , π

2)

.







(

−π

2 , π

2)

. Further, we have

tan (arctan x)







(

−π

2 , π

2)

. Further, we have

tan (arctan x) = x and







(

−π

2 , π

2)

. Further, we have

tan (arctan x) = x and arctan (tan x)







(

−π

2 , π

2)

. Further, we have

tan (arctan x) = x and arctan (tan x) = x for − π

2 < x <π

2







(

−π

2 , π

2)

. Further, we have


2 < x <π

2

Taking the derivative of the first of the formulas above, as we did to find thederivative of the ln function, gives

ddx

tan (arctan x)







(

−π

2 , π

2)

. Further, we have


2 < x <π

2


ddx

tan (arctan x) =(

1 + tan2 (arctan x)) d

dxarctan x







(

−π

2 , π

2)

. Further, we have


2 < x <π

2


ddx

tan (arctan x) =(


dxarctan x

=(

1 + x2) d

dxarctan x







(

−π

2 , π

2)

. Further, we have


2 < x <π

2


ddx

tan (arctan x) =(


dxarctan x

=(

1 + x2) d

dxarctan x =

ddx

x = 1







(

−π

2 , π

2)

. Further, we have


2 < x <π

2


ddx

tan (arctan x) =(


dxarctan x

=(

1 + x2) d

dxarctan x =

ddx

x = 1

Solving forddx

arctan x gives

ddx

arctan x







(

−π

2 , π

2)

. Further, we have


2 < x <π

2


ddx

tan (arctan x) =(


dxarctan x

=(

1 + x2) d

dxarctan x =

ddx

x = 1

Solving forddx

arctan x gives

ddx

arctan x =1

1 + x2





The arcsin Function

Recall that the graph of the sine function lookslike this.





The arcsin Function

Recall that the graph of the sine function lookslike this. From this graph we realize that thesine function is not one-to-one

π

2−π

2

1

−1





The arcsin Function

Recall that the graph of the sine function lookslike this. From this graph we realize that thesine function is not one-to-one, and so doesnot have an inverse function.

π

2−π

2

1

−1





The arcsin Function

Recall that the graph of the sine function lookslike this. From this graph we realize that thesine function is not one-to-one, and so doesnot have an inverse function. However,restricting the domain of the sine function,

π

2−π

2

1

−1





The arcsin Function

Recall that the graph of the sine function lookslike this. From this graph we realize that thesine function is not one-to-one, and so doesnot have an inverse function. However,restricting the domain of the sine function, asshown in the graph, to the interval

− π

2≤ x ≤ π

2

π

2−π

2

1

−1





The arcsin Function


− π

2≤ x ≤ π

2


π

2−π

2

1

−1





The arcsin Function


− π

2≤ x ≤ π

2


π

2−π

2

1

−1

So the inverse of the sine function is defined as

arcsin x = sin−1 x = the angle between − π

2 and π

2 whose sin is x





The arcsin Function


− π

2≤ x ≤ π

2


π

2−π

2

1

−1



2 and π

2 whose sin is x

The domain of inverse sine function, so defined, is





The arcsin Function


− π

2≤ x ≤ π

2


π

2−π

2

1

−1



2 and π

2 whose sin is x

The domain of inverse sine function, so defined, is [−1, 1] and its range is





The arcsin Function


− π

2≤ x ≤ π

2


π

2−π

2

1

−1



2 and π

2 whose sin is x

The domain of inverse sine function, so defined, is [−1, 1] and its range is[

−π

2 , π

2]

.





The Derivative of the arcsin Function

As with the arctan function, we have

sin (arcsin x)







sin (arcsin x) = x and







sin (arcsin x) = x and arcsin (sin x)







sin (arcsin x) = x and arcsin (sin x) = x for − π

2 ≤ x ≤ π

2








2 ≤ x ≤ π

2

Taking the derivative of the first of the formulas above, as we just did for thearctan, gives

ddx

sin (arcsin x)








2 ≤ x ≤ π

2


ddx

sin (arcsin x) = cos (arcsin x)ddx

arcsin x








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π

2 we have cos x ≥ 0








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π

2 we have cos x ≥ 0 , and using the basic Pythagorean

identity cos2 x + sin2 x = 1, gives cos x =√

1 − sin2 x.








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π



1 − sin2 x. So that

ddx

sin (arcsin x)








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π




ddx

sin (arcsin x) =√

1 − sin2 (arcsin x)ddx

arcsin x








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π




ddx

sin (arcsin x) =√


arcsin x =√

1 − x2 ddx

arcsin x








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π




ddx

sin (arcsin x) =√


arcsin x =√

1 − x2 ddx

arcsin x

Solving forddx

arcsin x gives

ddx

arcsin x








2 ≤ x ≤ π

2


ddx


arcsin x =ddx

x = 1

Now for −π

2 ≤ x ≤ π




ddx

sin (arcsin x) =√


arcsin x =√

1 − x2 ddx

arcsin x

Solving forddx

arcsin x gives

ddx

arcsin x =1√

1 − x2





Example 48 – Differentiating with Inverse Trig Functions

Find and simplify the indicated derivative(s) of each function.

(a) Find f ′(x) and f ′′(x) for f (x) =(

1 + x2) arctan x.

(b) Finddydx

for y = arcsin(√

1 − x2)

.






Using the Product Rule gives

f ′(x)







f ′(x) = 2x arctan x +(

1 + x2)

(

11 + x2

)








1 + x2)

(

11 + x2

)

= 2x arctan x + 1








1 + x2)

(

11 + x2

)

= 2x arctan x + 1

Taking the derivative of the expression above, using the Product Rule again,gives

f ′′(x)








1 + x2)

(

11 + x2

)

= 2x arctan x + 1


f ′′(x) = 2 arctan x + 2x(

11 + x2

)








1 + x2)

(

11 + x2

)

= 2x arctan x + 1


f ′′(x) = 2 arctan x + 2x(

11 + x2

)

= 2 arctan x +2x

1 + x2






Using the Chain Rule gives

dydx







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)

= − x|x|

√1 − x2







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)

= − x|x|

√1 − x2

Thus, if 0 ≤ x < 1, then

ddx

arcsin(√

1 − x2)

= − 1√1 − x2







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)

= − x|x|

√1 − x2


ddx

arcsin(√

1 − x2)

= − 1√1 − x2

But you can show that

ddx

arccos x = − 1√1 − x2







dydx

=

1√

1 −(√

1 − x2)2

(

12

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)

= − x|x|

√1 − x2


ddx

arcsin(√

1 − x2)

= − 1√1 − x2

But you can show that

ddx

arccos x = − 1√1 − x2

So is it true that arccos x = arcsin(√

1 − x2)

?


Documents

Math 112 Lecture 13: Differentiation -- Derivatives of ...€¦ · Derivatives of the Inverse Trigonometric Functions Derivative of sin Derivative of cos Using the Chain Rule Derivative