Derivative of Inverse Trigonometric Functions
The Derivatives of Inverse Trigonometric Functions Are as Follows: Derivative of Arcsin(X): D/Dx Arcsin(X) = 1 / √(1 - X^2) Derivative of Arccos(X): D/Dx...
The derivatives of inverse trigonometric functions are as follows:
Derivative of arcsin(x):
- d/dx arcsin(x) = 1 / √(1 - x^2)
Derivative of arccos(x):
- d/dx arccos(x) = -1 / √(1 - x^2)
Derivative of arctan(x):
- d/dx arctan(x) = 1 / (1 + x^2)
Derivative of arccsc(x):
- d/dx arccsc(x) = -1 / (|x|√(x^2 - 1))
Derivative of arcsec(x):
- d/dx arcsec(x) = 1 / (|x|√(x^2 - 1))
Derivative of arccot(x):
- d/dx arccot(x) = -1 / (1 + x^2)
These derivatives can be derived using the rules of differentiation and the trigonometric identities. Keep in mind that the domains of these functions are restricted to certain intervals to ensure that the derivatives exist. For example, the domain of arcsin(x) and arccos(x) is typically -1 ≤ x ≤ 1, and the domain of arctan(x) is all real numbers (-∞ < x < ∞).
What is Derivative of Inverse Trigonometric Functions?
The derivative of an inverse trigonometric function is a mathematical expression that describes how the rate of change of the inverse trigonometric function varies with respect to its input. Each of the six inverse trigonometric functions (arcsin, arccos, arctan, arccsc, arcsec, and arccot) has a specific derivative that can be expressed using basic calculus principles. Here are the derivatives of the most common inverse trigonometric functions:
Derivative of arcsin (Inverse Sine):
- d/dx [arcsin(x)] = 1 / √(1 - x^2)
Derivative of arccos (Inverse Cosine):
- d/dx [arccos(x)] = -1 / √(1 - x^2)
Derivative of arctan (Inverse Tangent):
- d/dx [arctan(x)] = 1 / (1 + x^2)
Derivative of arccsc (Inverse Cosecant):
- d/dx [arccsc(x)] = -1 / (|x| * √(x^2 - 1))
Derivative of arcsec (Inverse Secant):
- d/dx [arcsec(x)] = 1 / (|x| * √(x^2 - 1))
Derivative of arccot (Inverse Cotangent):
- d/dx [arccot(x)] = -1 / (1 + x^2)
These derivatives can be derived using the chain rule and trigonometric identities, and they are fundamental in calculus when dealing with functions involving inverse trigonometric functions. They are often used to find rates of change, gradients, and slopes in various mathematical and scientific applications.
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Examples of Derivative of Inverse Trigonometric Functions
Here are some examples of derivatives of inverse trigonometric functions:
|
Function |
Derivative |
|
sin⁻¹(x) |
1/√(1 - x²) |
|
cos⁻¹(x) |
-1/√(1 - x²) |
|
tan⁻¹(x) |
1/(1 + x²) |
|
csc⁻¹(x) |
`-1/( |
|
sec⁻¹(x) |
`1/( |
|
cot⁻¹(x) |
-1/(1 + x²) |
Here are some examples of how to use these derivatives:
Example 1: Find the derivative of the function f(x) = sin⁻¹(x) + 2cos⁻¹(x).
Solution: Using the derivatives of sin⁻¹(x) and cos⁻¹(x), we get:
f'(x) = d/dx(sin⁻¹(x) + 2cos⁻¹(x)) = 1/√(1 - x²) - 4/√(1 - x²) = (1 - 4)/√(1 - x²) = -3/√(1 - x²)
Example 2: Find the derivative of the function g(x) = x²sin⁻¹(x).
Solution: Using the product rule and the derivative of sin⁻¹(x), we get:
g'(x) = d/dx(x²sin⁻¹(x)) = x² d/dx(sin⁻¹(x)) + sin⁻¹(x) d/dx(x²) = x² * 1/√(1 - x²) + sin⁻¹(x) * 2x = (x² + 2xsin⁻¹(x))/√(1 - x²)
Example 3: Find the derivative of the function h(x) = tan⁻¹(x²) + sec⁻¹(x).
Solution: Using the derivatives of tan⁻¹(x) and sec⁻¹(x), we get:
h'(x) = d/dx(tan⁻¹(x²) + sec⁻¹(x)) = 1/(1 + x⁴) + 1/(|x|√(x² - 1))
Note that the derivative of sec⁻¹(x) is undefined at x = ±1. Therefore, the derivative of h(x) is also undefined at x = ±1.
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