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Derivatives of the Basic Trigonometric Functions Applying the Trig Function Derivative Rules Derivatives of the Inverse Trigonometric Functions

Lecture 13: 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 Lee Math 112 Lecture 13: Derivatives of Trigonometric Functions 1/30


Derivative of sin

Two Identities

In Example 31(c) we guessed that

d

dxsin x = cos x

from the graphs of sin and cos. This can be proved using the limitdefinition of the derivative together with some basic identities.



Derivative of sin

Two Identities


d

dxsin x = cos x


Sum and Difference Formulas for sin and cos



Derivative of sin

Two Identities


d

dxsin x = cos x



The sum and difference formulas for sin

sin (x ± y)



Derivative of sin

Two Identities


d

dxsin x = cos x




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



Derivative of sin

Two Identities


d

dxsin x = cos x





The corresponding formula for cos



Derivative of sin

Two Identities


d

dxsin x = cos x






cos (x ± y)



Derivative of sin

Two Identities


d

dxsin x = cos x






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



Derivative of sin

Derivative of sin

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



Derivative of sin

Derivative of sin


f ′(x)



Derivative of sin

Derivative of sin


f ′(x) = limh→0

sin (x + h) − sin x

h



Derivative of sin

Derivative of sin


f ′(x) = limh→0


h

= limh→0

sin x cos h + cos x sin h − sin x

h



Derivative of sin

Derivative of sin


f ′(x) = limh→0


h

= limh→0


h

= limh→0

sin x (cos h − 1) + cos x sin h

h



Derivative of sin

Derivative of sin


f ′(x) = limh→0


h

= limh→0


h

= limh→0


h

= sin x

(

limh→0

cos h − 1

h

)

+ cos x

(

limh→0

sin h

h

)



Derivative of sin

Derivative of sin


f ′(x) = limh→0


h

= limh→0


h

= limh→0


h

= sin x

(

limh→0

cos h − 1

h

)

+ cos x

(

limh→0

sin h

h

)

Earlier we used the Squeeze Theorem to proved that

limx→0

sin x

x



Derivative of sin

Derivative of sin


f ′(x) = limh→0


h

= limh→0


h

= limh→0


h

= sin x

(

limh→0

cos h − 1

h

)

+ cos x

(

limh→0

sin h

h

)

Earlier we used the Squeeze Theorem to proved that

limx→0

sin x

x= 1



Derivative of sin

Derivative of sin – continued

As in Example 13 we can show that

limh→0

cos h − 1

h



Derivative of sin



limh→0

cos h − 1

h= 0



Derivative of sin



limh→0

cos h − 1

h= 0

Thus

f ′(x) =d

dxsin x



Derivative of sin



limh→0

cos h − 1

h= 0

Thus

f ′(x) =d

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



Derivative of sin



limh→0

cos h − 1

h= 0

Thus

f ′(x) =d

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



Derivative of sin



limh→0

cos h − 1

h= 0

Thus

f ′(x) =d


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

d

dxcos x



Derivative of sin



limh→0

cos h − 1

h= 0

Thus

f ′(x) =d


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

d

dxcos x = − sin x



Derivative of cos Using the Chain Rule


Complementary Angle Identities

sin(

π

2 − x)






sin(

π

2 − x)

= cos x






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

=






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x

since x and π

2 − x are complementary angles in a right triangle.






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x

since x and π


Using the Chain Rule on the first of the identities above

d

dxcos x






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x

since x and π



d

dxcos x =

d

dxsin(

π

2 − x)






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x

since x and π



d

dxcos x =

d

dxsin(

π

2 − x)

= cos(

π

2 − x) d

dx

(

π

2 − x)






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x

since x and π



d

dxcos x =

d

dxsin(

π

2 − x)

= cos(

π

2 − x) d

dx

(

π

2 − x)

= sin x(−1)






sin(

π

2 − x)

= cos x

cos(

π

2 − x)

= sin x

since x and π



d

dxcos x =

d

dxsin(

π

2 − x)

= cos(

π

2 − x) d

dx

(

π

2 − x)

= sin x(−1) = − sin x




Two Basic Derivative Formulas

Derivatives of sin and cos

d

dxsin u =






d

dxsin u = cos u

du

dx






d

dxsin u = cos u

du

dxd

dxcos u =






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dx






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dx

Here the Chain Rule form of the derivative formulas are given, inwhich the variable u is a function of x.



Derivative of tan Using the Quotient Rule


Recall that

tan x





Recall that

tan x =sin x

cos x





Recall that

tan x =sin x

cos x

Thus, using the Quotient Rule gives

d

dxtan x





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x

=

(

d

dxsin x

)

cos x − sin x

(

d

dxcos x

)

(cos x)2





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x

=

(

d

dxsin x

)

cos x − sin x

(

d

dxcos x

)

(cos x)2

=(cos x) cos x − sin x (− sin x)

cos2 x





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x

=

(

d

dxsin x

)

cos x − sin x

(

d

dxcos x

)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x

=

(

d

dxsin x

)

cos x − sin x

(

d

dxcos x

)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x

=

(

d

dxsin x

)

cos x − sin x

(

d

dxcos x

)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x=

(

1

cos x

)2





Recall that

tan x =sin x

cos x


d

dxtan x =

d

dx

sin x

cos x

=

(

d

dxsin x

)

cos x − sin x

(

d

dxcos x

)

(cos x)2


cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x=

(

1

cos x

)2

= sec2 x




Derivative of tan Using the Quotient Rule – continued

An alternative way to simplify the previous expression is

d

dxtan x






d

dxtan x =

cos2 x + sin2 x

cos2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2

= 1 + tan2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2

= 1 + tan2 x

So thatd

dxtan x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2

= 1 + tan2 x

So thatd

dxtan x = sec2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2

= 1 + tan2 x

So thatd

dxtan x = sec2 x = 1 + tan2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2

= 1 + tan2 x

So thatd


The equality above can also be proved using the Pythagorean identity

1 + tan2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos x

)2

= 1 + tan2 x

So thatd



1 + tan2 x = sec2 x






d

dxtan x =

cos2 x + sin2 x

cos2 x=

cos2 x

cos2 x+

sin2 x

cos2 x

= 1 +

(

sin x

cos 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, butMaple and other symbolic differentiating programs use the 1 + tan2 xformula.



Derivatives the Six Trigonometric Functions


Derivatives Formulas for the Six Trigonometric Functions

d

dxsin u = cos u

du

dx






d

dxsin u = cos u

du

dxd

dxcos u






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dx






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dx






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dxd

dxsec u






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dxd

dxsec u = sec u tan u

du

dx






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dxd


du

dxd

dxcsc u






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dxd


du

dxd

dxcsc u = − csc u cot u

du

dx






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dxd


du

dxd


du

dxd

dxcot u






d

dxsin u = cos u

du

dxd

dxcos u = − sin u

du

dxd

dxtan u = sec2 u

du

dx=(

1 + tan2 u) du

dxd


du

dxd


du

dxd

dxcot u = − csc2 u

du

dx= −

(

1 + cot2 u) du

dx



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) Findds

dtfor 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 theProduct and Chain 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

ds

dt






ds

dt=

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

(sin t + cos t)2






ds

dt=


(sin t + cos t)2

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

(sin t + cos t)2






ds

dt=


(sin t + cos t)2


(sin t + cos t)2

=−(

sin2 t + cos2 t)

(sin t + cos t)2






ds

dt=


(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






ds

dt=


(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 derivativesinvolving trig functions, you sometimes need to use standardtrigonometric 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 function

q(t) = e−7t sin (24t)

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

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

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




Continuing Example 47

(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

d2q

dt2+ 14

dq

dt+ 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)]

This is the velocity of the mass at time t.




Solution: Example 47(a) – continued

Taking the derivative of the expression above for q′(t) gives

q′′(t)






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

e−7t[

−242 sin (24t) − 7(24) cos (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 [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 [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)]

This gives the acceleration of the 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 firsttime after 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 oppositedirection.





The graph of the function q(t) looks likethis.The amplitude of the motion decreasesfollowing an envelope given by thedecaying exponential function e−7t, asshown.

0.1 0.2 0.3 0.4





The graph of the function q(t) looks likethis.The amplitude of the motion decreasesfollowing an envelope given by thedecaying exponential function e−7t, asshown. As we saw in the last part, notonly does the amplitude decrease, but sodoes the velocity.

0.1 0.2 0.3 0.4





The graph of the function q(t) looks likethis.The amplitude of the motion decreasesfollowing an envelope given by thedecaying exponential function e−7t, asshown. As we saw in the last part, notonly does the amplitude decrease, but sodoes the velocity.Further, note that where the graph isconcave down, q′′(t) < 0, the mass isdecelerating. The velocity is getting lesspositive or more negative.

0.1 0.2 0.3 0.4





The graph of the function q(t) looks likethis.The amplitude of the motion decreasesfollowing an envelope given by thedecaying exponential function e−7t, asshown. As we saw in the last part, notonly does the amplitude decrease, but sodoes the velocity.Further, note that where the graph isconcave down, q′′(t) < 0, the mass isdecelerating. The velocity is getting lesspositive or more negative. And, wherethe graph is concave up, q′′(t) > 0, themass is accelerating. The velocity isgetting more positive or less negative.

0.1 0.2 0.3 0.4




Solution: Example 47(d)

The mass turns around whenv(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) =24

7 0.1 0.2 0.3 0.4






24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =24

7

⇒24t = arctan

(

24

7

)

+ kπ

0.1 0.2 0.3 0.4






24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =24

7

⇒24t = arctan

(

24

7

)

+ kπ

where k is any integer.

0.1 0.2 0.3 0.4






24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =24

7

⇒24t = arctan

(

24

7

)

+ kπ

where k is any integer. The first twopositive t values are t1 = 0.0536 andt2 = 0.1845.

0.1 0.2 0.3 0.4






24 cos (24t) − 7 sin (24t) = 0

⇒ tan (24t) =24

7

⇒24t = arctan

(

24

7

)

+ kπ

where k is any integer. The first twopositive t values are t1 = 0.0536 andt2 = 0.1845. At these values the tangentline to the graph of q(t) is horizontal, asshown.

0.1 0.2 0.3 0.4




Solution: Example 47(e)

From part (a) we have

d2q

dt2+ 14

dq

dt+ 625q






d2q

dt2+ 14

dq

dt+ 625q

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






d2q

dt2+ 14

dq

dt+ 625q

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

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






d2q

dt2+ 14

dq

dt+ 625q

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


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






d2q

dt2+ 14

dq

dt+ 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 Function

The arctan Function

Recall that the graph of the tangentfunction looks like this.



The arctan Function

The arctan Function

Recall that the graph of the tangentfunction looks like this. The tangentfunction is not one-to-one

π

2−π

2



The arctan Function

The arctan Function

Recall that the graph of the tangentfunction looks like this. The tangentfunction is not one-to-one, and so doesnot have an inverse function.

π

2−π

2



The arctan Function

The arctan Function

Recall that the graph of the tangentfunction looks like this. The tangentfunction is not one-to-one, and so doesnot have an inverse function. However,restricting the domain of the tangentfunction,

π

2−π

2



The arctan Function

The arctan Function

Recall that the graph of the tangentfunction looks like this. The tangentfunction is not one-to-one, and so doesnot have an inverse function. However,restricting the domain of the tangentfunction, as shown in the graph, to theinterval

− π

2< x <

π

2

π

2−π

2



The arctan Function

The arctan Function


− π

2< x <

π

2

gives a one-to-one function.

π

2−π

2



The arctan Function

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 arctan Function

The Derivative of the arctan Function

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



The arctan Function


With this definition of the inverse tangent function, its domain is allreal numbers and its range is



The arctan Function



(

−π

2 , π

2

)

.



The arctan Function



(

−π

2 , π

2

)

. Further, we have

tan (arctan x)



The arctan Function



(

−π

2 , π

2

)

. Further, we have

tan (arctan x) = x and



The arctan Function



(

−π

2 , π

2

)

. Further, we have

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



The arctan Function



(

−π

2 , π

2

)

. Further, we have

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

2 < x <π

2



The arctan Function



(

−π

2 , π

2

)

. Further, we have


2 < x <π

2

Taking the derivative of the first of the formulas above, as we did tofind the derivative of the ln function, gives

d

dxtan (arctan x)



The arctan Function



(

−π

2 , π

2

)

. Further, we have


2 < x <π

2


d

dxtan (arctan x) =

(

1 + tan2 (arctan x)) d

dxarctan x



The arctan Function



(

−π

2 , π

2

)

. Further, we have


2 < x <π

2


d

dxtan (arctan x) =

(


dxarctan x

=(

1 + x2) d

dxarctan x



The arctan Function



(

−π

2 , π

2

)

. Further, we have


2 < x <π

2


d

dxtan (arctan x) =

(


dxarctan x

=(

1 + x2) d

dxarctan x =

d

dxx = 1



The arctan Function



(

−π

2 , π

2

)

. Further, we have


2 < x <π

2


d

dxtan (arctan x) =

(


dxarctan x

=(

1 + x2) d

dxarctan x =

d

dxx = 1

Solving ford

dxarctan x gives

d

dxarctan x



The arctan Function



(

−π

2 , π

2

)

. Further, we have


2 < x <π

2


d

dxtan (arctan x) =

(


dxarctan x

=(

1 + x2) d

dxarctan x =

d

dxx = 1

Solving ford

dxarctan x gives

d

dxarctan x =

1

1 + x2



The arcsin Function

The arcsin Function

Recall that the graph of the sine functionlooks like this.



The arcsin Function

The arcsin Function

Recall that the graph of the sine functionlooks like this. From this graph werealize that the sine function is notone-to-one π

2−π

2

1

−1



The arcsin Function

The arcsin Function

Recall that the graph of the sine functionlooks like this. From this graph werealize that the sine function is notone-to-one, and so does not have aninverse function.

π

2−π

2

1

−1



The arcsin Function

The arcsin Function

Recall that the graph of the sine functionlooks like this. From this graph werealize that the sine function is notone-to-one, and so does not have aninverse function. However, restricting thedomain of the sine function,

π

2−π

2

1

−1



The arcsin Function

The arcsin Function

Recall that the graph of the sine functionlooks like this. From this graph werealize that the sine function is notone-to-one, and so does not have aninverse function. However, restricting thedomain of the sine function, as shown inthe graph, to the interval

− π

2≤ x ≤ π

2

π

2−π

2

1

−1



The arcsin Function

The arcsin Function


− π

2≤ x ≤ π

2


π

2−π

2

1

−1



The arcsin Function

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

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

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 itsrange is



The arcsin Function

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 itsrange is

[

−π

2 , π

2

]

.



The arcsin Function

The Derivative of the arcsin Function

As with the arctan function, we have

sin (arcsin x)



The arcsin Function



sin (arcsin x) = x and



The arcsin Function



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



The arcsin Function



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

2 ≤ x ≤ π

2



The arcsin Function




2 ≤ x ≤ π

2

Taking the derivative of the first of the formulas above, as we just didfor the arctan, gives

d

dxsin (arcsin x)



The arcsin Function




2 ≤ x ≤ π

2


d

dxsin (arcsin x) = cos (arcsin x)

d

dxarcsin x



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1

Now for −π

2 ≤ x ≤ π

2 we have cos x ≥ 0



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 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.



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1

Now for −π

2 ≤ x ≤ π



1 − sin2 x.So that

d

dxsin (arcsin x)



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1

Now for −π

2 ≤ x ≤ π




d

dxsin (arcsin x) =

√

1 − sin2 (arcsin x)d

dxarcsin x



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1

Now for −π

2 ≤ x ≤ π




d

dxsin (arcsin x) =

√


dxarcsin x =

√

1 − x2d

dxarcsin x



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1

Now for −π

2 ≤ x ≤ π




d

dxsin (arcsin x) =

√


dxarcsin x =

√

1 − x2d

dxarcsin x

Solving ford

dxarcsin x gives

d

dxarcsin x



The arcsin Function




2 ≤ x ≤ π

2


d


d

dxarcsin x =

d

dxx = 1

Now for −π

2 ≤ x ≤ π




d

dxsin (arcsin x) =

√


dxarcsin x =

√

1 − x2d

dxarcsin x

Solving ford

dxarcsin x gives

d

dxarcsin x =

1√1 − x2



The arcsin Function

Derivatives of the Inverse Trigonometric Functions

Derivative Formulas for the Two Inverse Trigonometric Functions

d

dxarctan u =

1

1 + u2

du

dx



The arcsin Function



d

dxarctan u =

1

1 + u2

du

dxd

dxarcsin u



The arcsin Function



d

dxarctan u =

1

1 + u2

du

dxd

dxarcsin u =

1√1 − u2

du

dx



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) Finddy

dxfor y = arcsin

(√1 − x2

)

.





Using the Product Rule gives

f ′(x)






f ′(x) = 2x arctan x +(

1 + x2)

(

1

1 + x2

)







1 + x2)

(

1

1 + x2

)

= 2x arctan x + 1







1 + x2)

(

1

1 + x2

)

= 2x arctan x + 1

Taking the derivative of the expression above, using the Product Ruleagain, gives

f ′′(x)







1 + x2)

(

1

1 + x2

)

= 2x arctan x + 1


f ′′(x) = 2 arctan x + 2x

(

1

1 + x2

)







1 + x2)

(

1

1 + x2

)

= 2x arctan x + 1


f ′′(x) = 2 arctan x + 2x

(

1

1 + x2

)

= 2 arctan x +2x

1 + x2





Using the Chain Rule gives

dy

dx






dy

dx=

1√

1 −(√

1 − x2)2

(

1

2

)

(

1 − x2)−1/2

(−2x)






dy

dx=

1√

1 −(√

1 − x2)2

(

1

2

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)






dy

dx=

1√

1 −(√

1 − x2)2

(

1

2

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)






dy

dx=

1√

1 −(√

1 − x2)2

(

1

2

)

(

1 − x2)−1/2

(−2x)

=

(

1√

1 − (1 − x2)

)

( −x√1 − x2

)

=

(

1√x2

)( −x√1 − x2

)

= − x

|x|√

1 − x2




Solution: Example 48(b) – continued

If 0 ≤ x < 1d

dxarcsin

(√

1 − x2)

= − 1√1 − x2

You can show that

d

dxarccos x = − 1√

1 − x2




Solution: Example 48(b) – continued

If 0 ≤ x < 1d

dxarcsin

(√

1 − x2)

= − 1√1 − x2

You can show that

d

dxarccos x = − 1√

1 − x2

So is it true that arccos x = arcsin(√

1 − x2)

?


math 112 lecture 13: derivatives of trigonometric functionspeople.okanagan.bc.ca › clee ›...

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