Derivative of Inverse Trigonometric functions

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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 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.

Examples of Derivative of Inverse Trigonometric Functions

Here are some examples of derivatives of inverse trigonometric functions:

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.

Proof of Derivative of Inverse Trigonometric Functions

To prove the derivatives of inverse trigonometric functions, we’ll start with the basic trigonometric identities and apply the rules of differentiation. Here, we’ll prove the derivatives of the arcsine (inverse sine), arccosine (inverse cosine), and arctangent (inverse tangent) functions.

Derivative of arcsin(x):

Let y = arcsin(x). This means sin(y) = x. Now, we’ll differentiate both sides of this equation with respect to x.

Using the chain rule on the left side (since y is a function of x), we get:

Now, solve for dy/dx:

But we know that sin(y) = x, and since sin^2(y) + cos^2(y) = 1 (from the Pythagorean trigonometric identity), we can express cos(y) as:

• cos(y) = √(1 – sin^2(y))

Substituting sin(y) = x into the equation above:

So, the derivative of arcsin(x) is:

Derivative of arccos(x):

Let y = arccos(x). This means cos(y) = x. Differentiating both sides:

Using the chain rule again:

Now, solve for dy/dx:

But we know that cos(y) = x, and sin^2(y) + cos^2(y) = 1, so:

Substituting sin(y) = √(1 – x^2) into the equation above:

• dy/dx = -1 / √(1 – x^2)

Derivative of arctan(x):

Let y = arctan(x). This means tan(y) = x. Differentiating both sides:

Using the chain rule:

Now, solve for dy/dx:

But we know that tan(y) = x, and sec^2(y) = 1 + tan^2(y), so:

Substituting sec^2(y) = 1 + x^2 into the equation above:

So, these are the derivatives of the inverse trigonometric functions:

• d/dx(arcsin(x)) = 1 / √(1 – x^2)
• d/dx(arccos(x)) = -1 / √(1 – x^2)
• d/dx(arctan(x)) = 1 / (1 + x^2)

These derivatives are fundamental for solving problems involving trigonometric functions in calculus.

Inverse Trigonometric Derivatives and Integrals

Here’s a table summarizing the derivatives and integrals of the most commonly used inverse trigonometric functions:

Please note that these formulas are valid within their respective domains and intervals, and the constant of integration (C) should be added when dealing with indefinite integrals.

Some Solved Problems on Derivative of Inverse Trigonometry

Here are some solved problems on the derivative of inverse trigonometry:

Problem 1: Find the derivative of f(x)=sin^{-1}(x).

Solution: We can use the following identity:

frac{d}{dx}left[sin^{-1}(x)right]=frac{1}{sqrt{1-x^2}}

This identity can be derived using the chain rule and the derivative of the sine function.

Therefore, the derivative of f(x)=sin^{-1}(x) is:

f'(x)=frac{1}{sqrt{1-x^2}}

Problem 2: Find the derivative of g(x)=cos^{-1}(x).

Solution: We can use the following identity:

frac{d}{dx}left[cos^{-1}(x)right]=-frac{1}{sqrt{1-x^2}}

This identity can be derived in a similar way to the identity for the derivative of sin^{-1}(x).

Therefore, the derivative of g(x)=cos^{-1}(x) is:

g'(x)=-frac{1}{sqrt{1-x^2}}

Problem 3: Find the derivative of h(x)=tan^{-1}(x).

Solution: We can use the following identity:

frac{d}{dx}left[tan^{-1}(x)right]=frac{1}{1+x^2}

This identity can be derived using the chain rule and the derivative of the tangent function.

Therefore, the derivative of h(x)=tan^{-1}(x) is:

h'(x)=frac{1}{1+x^2}

Problem 4: Find the derivative of the function `f(x)=2x^3+3sin^{-1}(x)$.

Solution: We can use the sum rule and the chain rule to find the derivative of this function.

f'(x)=6x^2+frac{3}{sqrt{1-x^2}}

Problem 5: Find the derivative of the function `g(x)=frac{tan^{-1}(x)}{x+1}$.

Solution: We can use the quotient rule and the chain rule to find the derivative of this function.

g'(x)=frac{frac{1}{1+x^2}(x+1)-tan^{-1}(x)}{(x+1)^2}

These are just a few examples of solved problems on the derivative of inverse trigonometry. There are many other problems that can be solved using the identities that we have presented here.

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