15 derivatives and integrals of inverse trigonometric functions

Derivatives and Integrals of the Inverse Trigonometric Functions

y = f(x)

y = f–1 (x)

The graphs of f and f–1 are symmetric diagonally.

y = x


y = f(x)

y = f–1 (x)


y = x


(a, b)

Assume f(x) is differentiable and (a, b) be a point on the graph of y = f(x).

y = f(x)

y = f–1 (x)


y = x


(a, b)

Assume f(x) is differentiable and (a, b) be a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a).

slope = f ’(a)

y = f(x)

y = f–1 (x)


y = x


(a, b)


(b, a)

slope = f ’(a)

The reflection of (a, b) is (b, a) on the graph of y = f–1(x).

y = f(x)

y = f–1 (x)


y = x


(a, b)


(b, a)

slope = f ’(a)

The reflection of (a, b) is (b, a) on the graph of y = f–1(x). The slope of the tangent line at (b, a) is (f–1)’(b).

slope = (f–1)’(b)



Assume f(x) is differentiable and (a, b) be a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a).The reflection of (a, b) is (b, a) on the graph of y = f–1(x). The slope of the tangent line at (b, a) is (f–1)’(b).By the symmetry the tangents at (a, b) and at (b, a) are the reflection of each other.

y = f(x)

y = f–1 (x)

y = x

(a, b)

(b, a)

slope = f ’(a)




Assume f(x) is differentiable and (a, b) be a point on the graph of y = f(x). Hence the slope at (a, b) is f ’(a).The reflection of (a, b) is (b, a) on the graph of y = f–1(x). The slope of the tangent line at (b, a) is (f–1)’(b).By the symmetry the tangents at (a, b) and at (b, a) are the reflection of each other. Diagonally-symmetric lines have reciprocal slopes (why?)

y = f(x)

y = f–1 (x)

y = x

(a, b)

(b, a)

slope = f ’(a)





y = f(x)

y = f–1 (x)

y = x

(a, b)

(b, a)

slope = f ’(a)


Hence of (f–1)’(b) = 1f ’(a)




y = f(x)

y = f–1 (x)

y = x

(a, b)

(b, a)

slope = f ’(a)


Hence of (f–1)’(b) = 1 = 1f ’(f–1(b))f ’(a)

Suppose f and g is a pair of inverse functions, then (f o g)(x) = x.



Differentiate both sides with respect to x and uses the chain rule:

[(f o g)(x)]' = x'




[(f o g)(x)]' = x'

f '(g(x)) * g'(x) = 1




[(f o g)(x)]' = x'

f '(g(x)) * g'(x) = 1

or

g'(x) = 1f '(g(x))




[(f o g)(x)]' = x'

f '(g(x)) * g'(x) = 1

or

g'(x) = 1f '(g(x))

Set f = sin(x) and g = arcsin(x)




[(f o g)(x)]' = x'

f '(g(x)) * g'(x) = 1

or

g'(x) = 1f '(g(x))

Set f = sin(x) and g = arcsin(x) we obtain

[arcsin(x)]' =

1 dsin(y)


dy y=arcsin(x)



[(f o g)(x)]' = x'

f '(g(x)) * g'(x) = 1

or

g'(x) = 1f '(g(x))

Set f = sin(x) and g = arcsin(x) we obtain

[arcsin(x)]' =

1 dsin(y)

1cos(arcsin(x))

=


dy y=arcsin(x)

[arcsin(x)]'

1cos(arcsin(x)) =


θ=arcsin(x)

x1

[arcsin(x)]'

1cos(arcsin(x)) =


θ=arcsin(x)

x1

1 – x2

[arcsin(x)]'

1cos(arcsin(x)) =


θ=arcsin(x)

x1

1 – x2

[arcsin(x)]'

1cos(arcsin(x)) =

11 – x2

[arcsin(x)]' =


θ=arcsin(x)

x1

1 – x2

[arcsin(x)]'

1cos(arcsin(x)) =

11 – x2

[arcsin(x)]' =


–1 1

–π/2

π/2

x

y=arcsin(x)

θ=arcsin(x)

x1

1 – x2

[arcsin(x)]'

1cos(arcsin(x)) =

11 – x2

[arcsin(x)]' = We use the same technique to obtain the derivatives of the other inverse trig-functions.


–1 1

–π/2

π/2

x

y=arcsin(x)

θ=arcsin(x)

x1

1 – x2

[arcsin(x)]'

1cos(arcsin(x)) =

11 – x2

[arcsin(x)]' = We use the same technique to obtain the derivatives of the other inverse trig-functions.


–1 1

–π/2

π/2

x

y=arcsin(x)

We display the graphs of each inverse–trig and list each of their derivatives below.

Derivatives of the Inverse Trig–Functions1

1 – x2[sin–1(x)] ' =

y =sin–1(x)

Derivatives of the Inverse Trig–Functions

[cos–1(x)]' = 11 – x2

[sin–1(x)] ' = –11 – x2

y =sin–1(x)y =cos–1(x)


[cos–1(x)]' =

1 1 + x2

[tan–1(x)]' =

11 – x2

[sin–1(x)] ' = –11 – x2

y =tan–1(x)



[cos–1(x)]' =

1 1 + x2

[tan–1(x)]' = –1 1 + x2[cot–1(x)]' =

11 – x2

[sin–1(x)] ' = –11 – x2

y =tan–1(x)


y =cot–1(x)


[cos–1(x)]' =

1 1 + x2

[tan–1(x)]' =

|x|x2 – 1[sec–1(x)]' =

–1 1 + x2[cot–1(x)]' =

y =sec–1(x)

11 – x2

[sin–1(x)] ' = –11 – x2

1

y =tan–1(x)


y =cot–1(x)


[cos–1(x)]' =

1 1 + x2

[tan–1(x)]' =

|x|x2 – 1[sec–1(x)]' =

–1 1 + x2[cot–1(x)]' =

–1|x|x2 – 1

[csc–1(x)]' =

y =sec–1(x) y =csc–1(x)

11 – x2

[sin–1(x)] ' = –11 – x2

1

y =tan–1(x)


y =cot–1(x)


u'1 – u2[sin–1(u)]' = –u'

1 – u2[cos–1 (u)]' =

u' 1 + u2[tan–1(u)]' =

u'|u|u2 – 1

[sec–1(u)]' =

–u' 1 + u2[cot–1(u)]' =

–u'|u|u2 – 1

[csc–1(u)]' =

dsin-1(u)dx

=

11 – u2

dudx

–11 – u2

dudx

11 + u2

dudx

1|u|u2 – 1

dudx

dcos-1(u)dx

=

dtan-1(u)dx

=

dsec-1(u)dx

=

–11 + u2

dudx

dtan-1(u)dx

=

–1|u|u2 – 1

dudx

dcsc-1(u)dx

=

Below are the chain–rule versions where u = u(x).


–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =

We’ll use the following formulas for the next example.


b. cos–1(ex )

c. sec–1(ln(x))]

Example A. Find the following derivatives.

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =



b. cos–1(ex )

2

c. sec–1(ln(x))]


Set u = x3, a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =



b. cos–1(ex )

2

c. sec–1(ln(x))]


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =



b. cos–1(ex )

2

c. sec–1(ln(x))]


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =



b. cos–1(ex ) Set u = ex,

2

c. sec–1(ln(x))]


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =



b. cos–1(ex ) Set u = ex, so [cos-1(ex )]' =

2

2

1 – (ex )2

–(ex )'2

2

c. sec–1(ln(x))]


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =




2

2

1 – (ex )2

–(ex )'2

2

=–2xex

1 – e2x 2

2

c. sec–1(ln(x))]


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =




2

2

1 – (ex )2

–(ex )'2

2

=–2xex

1 – e2x 2

2

c. sec–1(ln(x))] Set u = ln(x), so [sec–1(ln(x)]'

=


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =




2

2

1 – (ex )2

–(ex )'2

2

=–2xex

1 – e2x 2

2


= 1/x

|ln(x)|ln2(x) – 1


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =




2

2

1 – (ex )2

–(ex )'2

2

=–2xex

1 – e2x 2

2


= 1/x

|ln(x)|ln2(x) – 1= 1

x|ln(x)|ln2(x) – 1


Set u = x3, so [tan–1(x3)]' = (x3)'1 + (x3)2 =

3x2

1 + x6

a. tan–1(x3)

–u'

1 – u2[cos–1 (u)]' = u'

1 + u2[tan–1(u)]' = u'

|u|u2 – 1[sec–1(u)]' =



= sin-1(u) + C 1 – u2

du

Expressing the relations in integrals:

∫


= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫


= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2


= sin-1(u) + C 1 – u2

du

|u|u2 – 1


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2

= sec-1(u) + C

∫ du


= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2

= sec-1(u) + C

∫

Example B. Find the integral ∫ dx9 + 4x2

|u|u2 – 1 du


= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2

= sec-1(u) + C

∫

Match the form of the integral to the one for tan-1(u).

|u|u2 – 1 du



= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2

= sec-1(u) + C

∫

Match the form of the integral to the one for tan-1(u). Write 9 + 4x2 = 9 (1 + x2)

49

|u|u2 – 1 du



= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2

= sec-1(u) + C

∫

Match the form of the integral to the one for tan-1(u). Write 9 + 4x2 = 9 (1 + x2) = 9 [1 + ( x)2] 2

3 49

|u|u2 – 1 du



= sin-1(u) + C 1 – u2

du


∫

= cos-1(u) + C 1 – u2

- du∫

= tan-1(u) + C

du∫ 1 + u2

= sec-1(u) + C

∫

Match the form of the integral to the one for tan-1(u). Write 9 + 4x2 = 9 (1 + x2) = 9 [1 + ( x)2] 2

3 49

Hence dx9 + 4x2 ∫ = dx

1 + ( x)2 ∫ 19 2

3

|u|u2 – 1 du



dx1 + ( x)2 ∫

19 2

3

substitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x

substitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x du

dx =

23

substitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x du

dx =

23

So dx =

32

du

substitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x du

dx =

23

So dx =

32

du= ∫

19

11 + u2

32

dusubstitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x du

dx =

23

So dx =

32

du= ∫

19

11 + u2

32

du

= ∫ 16

11 + u2

du

substitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x du

dx =

23

So dx =

32

du= ∫

19

11 + u2

32

du

= ∫ 16

11 + u2

du

= tan-1(u) + C 16

substitution method


Set u =

dx1 + ( x)2 ∫

19 2

3

23 x du

dx =

23

So dx =

32

du= ∫

19

11 + u2

32

du

= ∫ 16

11 + u2

du

= tan-1(u) + C 16

= tan-1( x) + C 16

23

substitution method


ex

∫Example C. Find the definite integral 1 – e2x

dx0

ln(1/2)


ex

∫Example C. Find the definite integral 1 – e2x

dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)


ex

∫

substitution method

Example C. Find the definite integral 1 – e2x

dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)


Set u =

ex

∫

substitution method


dx0

ln(1/2)

ex

ex

∫1 – e2x

dx0

ln(1/2)


Set u =

ex

∫

dudx =

substitution method


dx0

ln(1/2)

ex ex

ex

∫1 – e2x

dx0

ln(1/2)


Set u =

ex

∫

dudx =

So dx =

du/ex

substitution method


dx0

ln(1/2)

ex ex

ex

∫1 – e2x

dx0

ln(1/2)


Set u =

ex

∫

dudx =

So dx =

du/ex

substitution method


dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

ex

∫1 – e2x

dx0

ln(1/2)


Set u =

ex

∫

dudx =

So dx =

du/ex

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du ex


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du1

1/2 ex


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du1

1/2 ex

= ∫1 – u2

du1

1/2


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du1

1/2 ex

= ∫1 – u2

du1

1/2

= sin-1(u) |1/2

1


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du1

1/2 ex

= ∫1 – u2

du1

1/2

= sin-1(u) |1/2

1

= sin-1(1) – sin-1(1/2)


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du1

1/2 ex

= ∫1 – u2

du1

1/2

= sin-1(u) |1/2

1

= sin-1(1) – sin-1(1/2) = π/2 – π/6


Set u =

ex

∫

dudx =

So dx =

du/ex=

substitution method


dx0

ln(1/2)

ex

∫1 – e2x

dx0

ln(1/2)

ex ex

for x = ln(1/2) u = 1/2

x = 0 u = 1

ex

∫1 – u2

du1

1/2 ex

= ∫1 – u2

du1

1/2

= sin-1(u) |1/2

1

= sin-1(1) – sin-1(1/2) = π/2 – π/6 = π/3

Lastly, we have the hyperbolic trigonometricfunctions.


Lastly, we have the hyperbolic trigonometricfunctions. These functions are made from the exponential functions with relations amongst them are similar to the trig-family.




We define the hyperbolic sine as

sinh(x) = (pronounced as "sinsh of x")ex – e-x

2




sinh(x) = (pronounced as "sinsh of x")

We define the hyperbolic cosine as

cosh(x) = (pronounced as "cosh of x")

ex – e-x

2

ex + e-x

2




sinh(x) = (pronounced as "sinsh of x")

We define the hyperbolic cosine as

cosh(x) = (pronounced as "cosh of x")

ex – e-x

2

ex + e-x

2We define the hyperbolic tangent, cotangent,secant and cosecant as in the trig-family.


The hyperbolic tangent:

tanh(x) = sinh(x)cosh(x) =

ex – e-x

ex + e-x

The hyperbolic cotangent:

coth(x) = cosh(x)sinh(x) = ex – e-x

ex + e-x

The hyperbolic secant:

sech(x) = 1

cosh(x) = ex + e-x

2

The hyperbolic cosecant:

csch(x) = 1

sinh(x) = ex – e-x

2

As for the trig-family, we've the hyperbolic-trig hexagram to help us with their relations.


Hyperbolic Trig Hexagram



sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1

co-side


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1

Starting from any position, take two steps without turning, we have

I =

II III


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1


I =

II III

For example, staring at cosh(x), go to sinh(x) then to tanh(x),


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1


I =

II III

For example, staring at cosh(x), go to sinh(x) then to tanh(x), we get the relation

cosh(x) =

sinh(x) tanh(x)


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1


I =

II III

For example, staring at cosh(x), go to sinh(x) then to tanh(x), we get the relation

cosh(x) = , similarly sech(x) =

sinh(x) tanh(x)

csch(x) coth(x)


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1

Square-differenceRelations:


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1

Square-differenceRelations:The three upside down triangles gives the sq-differernce relations.


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1

Square-differenceRelations:The three upside down triangles gives the sq-differernce relations. Difference of the squares on top is the square of the bottom one.


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1


cosh2(x) – sinh2(x) = 1


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1



coth2(x) – 1 = csch2(x)


sinh(x)cosh(x)

coth(x)

csch(x)

tanh(x)

sech(x)

1



coth2(x) – 1 = csch2(x)

1 – tanh2(x) = sech2(x)


Hyperbolic trig-functions show up in engineering.


Hyperbolic trig-functions show up in engineering. Specifically the graph y = cosh(x) gives the shape of a hanging cable.


Graph of y = cosh(x)

Hyperbolic trig-functions show up in engineering. Specifically the graph y = cosh(x) gives the shape of a hanging cable.

(0, 1)


The derivatives of the hyperbolic trig-functionsare similar, but not the same as, the trig-family.


The derivatives of the hyperbolic trig-functionsare similar, but not the same as, the trig-family.One may easily check that:

[sinh(x)]' = cosh(x)

[cosh(x)]' = sinh(x)





[tanh(x)]' = sech2(x)

[coth(x)]' = -csch2(x)







[sech(x)]' = -sech(x)tanh(x)

[csch(x)]' = -csch(x)coth(x)

Frank Ma2006







[sech(x)]' = -sech(x)tanh(x)

[csch(x)]' = -csch(x)coth(x)

Frank Ma2006

HW. Write down the chain–rule versions of the derivatives of the hyperbolic trig-functions.

15 derivatives and integrals of inverse trigonometric functions

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