15 derivatives and integrals of inverse trigonometric functions
Transcript of 15 derivatives and integrals of inverse trigonometric functions
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Derivatives and Integrals of the Inverse Trigonometric Functions
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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
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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).
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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)
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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).
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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)
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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)
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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)
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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)
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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)
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Suppose f and g is a pair of inverse functions, then (f o g)(x) = x.
Derivatives and Integrals of the Inverse Trigonometric Functions
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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
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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
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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
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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
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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)
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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)
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[arcsin(x)]'
1cos(arcsin(x)) =
Derivatives and Integrals of the Inverse Trigonometric Functions
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θ=arcsin(x)
x1
[arcsin(x)]'
1cos(arcsin(x)) =
Derivatives and Integrals of the Inverse Trigonometric Functions
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θ=arcsin(x)
x1
1 – x2
[arcsin(x)]'
1cos(arcsin(x)) =
Derivatives and Integrals of the Inverse Trigonometric Functions
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θ=arcsin(x)
x1
1 – x2
[arcsin(x)]'
1cos(arcsin(x)) =
11 – x2
[arcsin(x)]' =
Derivatives and Integrals of the Inverse Trigonometric Functions
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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)
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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)
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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.
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Derivatives of the Inverse Trig–Functions1
1 – x2[sin–1(x)] ' =
y =sin–1(x)
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Derivatives of the Inverse Trig–Functions
[cos–1(x)]' = 11 – x2
[sin–1(x)] ' = –11 – x2
y =sin–1(x)y =cos–1(x)
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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)
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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)
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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)
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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)
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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).
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
![Page 42: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/42.jpg)
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.
![Page 43: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/43.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
= sin-1(u) + C 1 – u2
du
Expressing the relations in integrals:
∫
![Page 44: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/44.jpg)
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∫
![Page 45: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/45.jpg)
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
![Page 46: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/46.jpg)
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
![Page 47: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/47.jpg)
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
![Page 48: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/48.jpg)
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
![Page 49: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/49.jpg)
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
![Page 50: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/50.jpg)
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
![Page 51: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/51.jpg)
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
![Page 52: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/52.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
dx1 + ( x)2 ∫
19 2
3
substitution method
![Page 53: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/53.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
dx1 + ( x)2 ∫
19 2
3
23 x
substitution method
![Page 54: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/54.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
Set u =
dx1 + ( x)2 ∫
19 2
3
23 x du
dx =
23
substitution method
![Page 55: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/55.jpg)
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
![Page 56: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/56.jpg)
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
![Page 57: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/57.jpg)
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
![Page 58: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/58.jpg)
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
![Page 59: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/59.jpg)
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
![Page 60: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/60.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
ex
∫Example C. Find the definite integral 1 – e2x
dx0
ln(1/2)
![Page 61: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/61.jpg)
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)
![Page 62: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/62.jpg)
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)
![Page 63: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/63.jpg)
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)
![Page 64: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/64.jpg)
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)
![Page 65: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/65.jpg)
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)
![Page 66: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/66.jpg)
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)
![Page 67: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/67.jpg)
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
![Page 68: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/68.jpg)
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
![Page 69: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/69.jpg)
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
![Page 70: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/70.jpg)
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
![Page 71: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/71.jpg)
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
![Page 72: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/72.jpg)
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
![Page 73: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/73.jpg)
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)
![Page 74: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/74.jpg)
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
![Page 75: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/75.jpg)
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
![Page 76: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/76.jpg)
Lastly, we have the hyperbolic trigonometricfunctions.
Derivatives and Integrals of the Inverse Trigonometric Functions
![Page 77: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/77.jpg)
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
![Page 78: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/78.jpg)
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
![Page 79: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/79.jpg)
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
![Page 80: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/80.jpg)
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.
![Page 81: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/81.jpg)
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
![Page 82: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/82.jpg)
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
![Page 83: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/83.jpg)
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
![Page 84: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/84.jpg)
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
![Page 85: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/85.jpg)
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),
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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)
![Page 87: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/87.jpg)
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)
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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
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Derivatives and Integrals of the Inverse Trigonometric Functions
sinh(x)cosh(x)
coth(x)
csch(x)
tanh(x)
sech(x)
1
Square-differenceRelations:
![Page 90: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/90.jpg)
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.
![Page 91: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/91.jpg)
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.
![Page 92: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/92.jpg)
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
![Page 93: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/93.jpg)
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)
![Page 94: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/94.jpg)
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)
![Page 95: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/95.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
Hyperbolic trig-functions show up in engineering.
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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.
![Page 97: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/97.jpg)
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)
![Page 98: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/98.jpg)
Derivatives and Integrals of the Inverse Trigonometric Functions
The derivatives of the hyperbolic trig-functionsare similar, but not the same as, the trig-family.
![Page 99: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/99.jpg)
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)
![Page 100: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/100.jpg)
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)
![Page 101: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/101.jpg)
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
![Page 102: 15 derivatives and integrals of inverse trigonometric functions](https://reader035.fdocuments.in/reader035/viewer/2022081502/5562d775d8b42a6c498b5150/html5/thumbnails/102.jpg)
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.