15 derivatives and integrals of inverse trigonometric functions
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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
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
Derivatives and Integrals of the Inverse Trigonometric Functions
(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)
The graphs of f and f–1 are symmetric diagonally.
y = x
Derivatives and Integrals of the Inverse Trigonometric Functions
(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)
The graphs of f and f–1 are symmetric diagonally.
y = x
Derivatives and Integrals of the Inverse Trigonometric Functions
(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).
(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)
The graphs of f and f–1 are symmetric diagonally.
y = x
Derivatives and Integrals of the Inverse Trigonometric Functions
(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).
(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)
The graphs of f and f–1 are symmetric diagonally.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
slope = (f–1)’(b)
The graphs of f and f–1 are symmetric diagonally.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
slope = (f–1)’(b)
The graphs of f and f–1 are symmetric diagonally.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
slope = (f–1)’(b)
Hence of (f–1)’(b) = 1f ’(a)
The graphs of f and f–1 are symmetric diagonally.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
slope = (f–1)’(b)
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.
Derivatives and Integrals of the Inverse Trigonometric Functions
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'
Derivatives and Integrals of the Inverse Trigonometric Functions
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 '(g(x)) * g'(x) = 1
Derivatives and Integrals of the Inverse Trigonometric Functions
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 '(g(x)) * g'(x) = 1
or
g'(x) = 1f '(g(x))
Derivatives and Integrals of the Inverse Trigonometric Functions
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 '(g(x)) * g'(x) = 1
or
g'(x) = 1f '(g(x))
Set f = sin(x) and g = arcsin(x)
Derivatives and Integrals of the Inverse Trigonometric Functions
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 '(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)
Derivatives and Integrals of the Inverse Trigonometric Functions
dy y=arcsin(x)
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 '(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))
=
Derivatives and Integrals of the Inverse Trigonometric Functions
dy y=arcsin(x)
[arcsin(x)]'
1cos(arcsin(x)) =
Derivatives and Integrals of the Inverse Trigonometric Functions
θ=arcsin(x)
x1
[arcsin(x)]'
1cos(arcsin(x)) =
Derivatives and Integrals of the Inverse Trigonometric Functions
θ=arcsin(x)
x1
1 – x2
[arcsin(x)]'
1cos(arcsin(x)) =
Derivatives and Integrals of the Inverse Trigonometric Functions
θ=arcsin(x)
x1
1 – x2
[arcsin(x)]'
1cos(arcsin(x)) =
11 – x2
[arcsin(x)]' =
Derivatives and Integrals of the Inverse Trigonometric Functions
θ=arcsin(x)
x1
1 – x2
[arcsin(x)]'
1cos(arcsin(x)) =
11 – x2
[arcsin(x)]' =
Derivatives and Integrals of the Inverse Trigonometric 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.
Derivatives and Integrals of the Inverse Trigonometric 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.
Derivatives and Integrals of the Inverse Trigonometric 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)
Derivatives of the Inverse Trig–Functions
[cos–1(x)]' =
1 1 + x2
[tan–1(x)]' =
11 – x2
[sin–1(x)] ' = –11 – x2
y =tan–1(x)
y =sin–1(x)y =cos–1(x)
Derivatives of the Inverse Trig–Functions
[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 =sin–1(x)y =cos–1(x)
y =cot–1(x)
Derivatives of the Inverse Trig–Functions
[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 =sin–1(x)y =cos–1(x)
y =cot–1(x)
Derivatives of the Inverse Trig–Functions
[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 =sin–1(x)y =cos–1(x)
y =cot–1(x)
Derivatives of the Inverse Trig–Functions
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).
Derivatives and Integrals of the Inverse Trigonometric Functions
–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.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex )
2
c. sec–1(ln(x))]
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex )
2
c. sec–1(ln(x))]
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex )
2
c. sec–1(ln(x))]
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex ) Set u = ex,
2
c. sec–1(ln(x))]
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex ) Set u = ex, so [cos-1(ex )]' =
2
2
1 – (ex )2
–(ex )'2
2
c. sec–1(ln(x))]
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex ) Set u = ex, so [cos-1(ex )]' =
2
2
1 – (ex )2
–(ex )'2
2
=–2xex
1 – e2x 2
2
c. sec–1(ln(x))]
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex ) Set u = ex, so [cos-1(ex )]' =
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)]'
=
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex ) Set u = ex, so [cos-1(ex )]' =
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)]'
= 1/x
|ln(x)|ln2(x) – 1
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
b. cos–1(ex ) Set u = ex, so [cos-1(ex )]' =
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)]'
= 1/x
|ln(x)|ln2(x) – 1= 1
x|ln(x)|ln2(x) – 1
Example A. Find the following derivatives.
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)]' =
We’ll use the following formulas for the next example.
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= cos-1(u) + C 1 – u2
- du∫
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= cos-1(u) + C 1 – u2
- du∫
= tan-1(u) + C
du∫ 1 + u2
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
|u|u2 – 1
Expressing the relations in integrals:
∫
= cos-1(u) + C 1 – u2
- du∫
= tan-1(u) + C
du∫ 1 + u2
= sec-1(u) + C
∫ du
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= 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
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= 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
Example B. Find the integral ∫ dx9 + 4x2
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= 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
Example B. Find the integral ∫ dx9 + 4x2
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= 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
Example B. Find the integral ∫ dx9 + 4x2
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
= 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
Example B. Find the integral ∫ dx9 + 4x2
Derivatives and Integrals of the Inverse Trigonometric Functions
dx1 + ( x)2 ∫
19 2
3
substitution method
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
dx1 + ( x)2 ∫
19 2
3
23 x
substitution method
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
dx1 + ( x)2 ∫
19 2
3
23 x du
dx =
23
substitution method
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
dx1 + ( x)2 ∫
19 2
3
23 x du
dx =
23
So dx =
32
du
substitution method
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
dx1 + ( x)2 ∫
19 2
3
23 x du
dx =
23
So dx =
32
du= ∫
19
11 + u2
32
dusubstitution method
Derivatives and Integrals of the Inverse Trigonometric Functions
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
Derivatives and Integrals of the Inverse Trigonometric Functions
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
Derivatives and Integrals of the Inverse Trigonometric Functions
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
Derivatives and Integrals of the Inverse Trigonometric Functions
ex
∫Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
ex
∫Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
ex
∫1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
ex
∫
substitution method
Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
ex
∫1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
substitution method
Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
ex
ex
∫1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
substitution method
Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
ex ex
ex
∫1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex
substitution method
Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
ex ex
ex
∫1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex
substitution method
Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
ex ex
for x = ln(1/2) u = 1/2
ex
∫1 – e2x
dx0
ln(1/2)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
ex
∫
dudx =
So dx =
du/ex=
substitution method
Example C. Find the definite integral 1 – e2x
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.
Derivatives and Integrals of the Inverse Trigonometric Functions
Lastly, we have the hyperbolic trigonometricfunctions. These functions are made from the exponential functions with relations amongst them are similar to the trig-family.
Derivatives and Integrals of the Inverse Trigonometric Functions
Lastly, we have the hyperbolic trigonometricfunctions. These functions are made from the exponential functions with relations amongst them are similar to the trig-family.
Derivatives and Integrals of the Inverse Trigonometric Functions
We define the hyperbolic sine as
sinh(x) = (pronounced as "sinsh of x")ex – e-x
2
Lastly, we have the hyperbolic trigonometricfunctions. These functions are made from the exponential functions with relations amongst them are similar to the trig-family.
Derivatives and Integrals of the Inverse Trigonometric Functions
We define the hyperbolic sine as
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
Lastly, we have the hyperbolic trigonometricfunctions. These functions are made from the exponential functions with relations amongst them are similar to the trig-family.
Derivatives and Integrals of the Inverse Trigonometric Functions
We define the hyperbolic sine as
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.
Derivatives and Integrals of the Inverse Trigonometric Functions
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.
Derivatives and Integrals of the Inverse Trigonometric Functions
Hyperbolic Trig Hexagram
As for the trig-family, we've the hyperbolic-trig hexagram to help us with their relations.
Derivatives and Integrals of the Inverse Trigonometric Functions
sinh(x)cosh(x)
coth(x)
csch(x)
tanh(x)
sech(x)
1
co-side
Derivatives and Integrals of the Inverse Trigonometric Functions
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
Derivatives and Integrals of the Inverse Trigonometric Functions
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
For example, staring at cosh(x), go to sinh(x) then to tanh(x),
Derivatives and Integrals of the Inverse Trigonometric Functions
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
For example, staring at cosh(x), go to sinh(x) then to tanh(x), we get the relation
cosh(x) =
sinh(x) tanh(x)
Derivatives and Integrals of the Inverse Trigonometric Functions
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
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)
As for the trig-family, we've the hyperbolic-trig hexagram to help us with their relations.
Derivatives and Integrals of the Inverse Trigonometric Functions
Derivatives and Integrals of the Inverse Trigonometric Functions
sinh(x)cosh(x)
coth(x)
csch(x)
tanh(x)
sech(x)
1
Square-differenceRelations:
Derivatives and Integrals of the Inverse Trigonometric Functions
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.
Derivatives and Integrals of the Inverse Trigonometric Functions
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.
Derivatives and Integrals of the Inverse Trigonometric Functions
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.
cosh2(x) – sinh2(x) = 1
Derivatives and Integrals of the Inverse Trigonometric Functions
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.
cosh2(x) – sinh2(x) = 1
coth2(x) – 1 = csch2(x)
Derivatives and Integrals of the Inverse Trigonometric Functions
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.
cosh2(x) – sinh2(x) = 1
coth2(x) – 1 = csch2(x)
1 – tanh2(x) = sech2(x)
Derivatives and Integrals of the Inverse Trigonometric Functions
Hyperbolic trig-functions show up in engineering.
Derivatives and Integrals of the Inverse Trigonometric Functions
Hyperbolic trig-functions show up in engineering. Specifically the graph y = cosh(x) gives the shape of a hanging cable.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
Derivatives and Integrals of the Inverse Trigonometric Functions
The derivatives of the hyperbolic trig-functionsare similar, but not the same as, the trig-family.
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
Derivatives and Integrals of the Inverse Trigonometric Functions
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)
Derivatives and Integrals of the Inverse Trigonometric Functions
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
Derivatives and Integrals of the Inverse Trigonometric Functions
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
HW. Write down the chain–rule versions of the derivatives of the hyperbolic trig-functions.
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