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Derivative of Cot Inverse

The derivative of cot inverse, also known as the derivative of arccot, gives the value of the differentiation of cot inverse x which is the rate change of the change in the function cot inverse with respect to the variable. The derivative of cot inverse x is equal to -1/(1 + x²). We can evaluate the differentiation of cot inverse x using different methods of derivatives such as the first principle of derivatives, that is, the definition of limits and the implicit differentiation method. Hence, we can say that the rate of change of cot inverse is given by -1/(1 + x²).

Further, in this article, we will explore the derivative of cot inverse and derive its formula using various differentiation methods. We will also solve various examples related to the concept of differentiation of cot inverse for a better understanding.

1.	What is Derivative of Cot Inverse?
2.	Derivative of Cot Inverse x Formula
3.	Derivative of Cot Inverse x Proof
4.	Differentiation of Cot Inverse x By First Principle
5.	FAQs on Derivative of Cot Inverse

What is Derivative of Cot Inverse?

The derivative of cot inverse is equal to -1/(1 + x²) which is mathematically written as d(cot^-1)/dx = -1/(1 + x²) = d(arccot)/dx. We can evaluate the cot inverse derivative using various differentiation methods including the first principle of differentiation and implicit differentiation method. As we know that differentiation of a function is a process of finding the rate of change of a function with respect to the change in the variable. We know that the derivative of tan inverse x is equal to 1/(1 + x²), therefore the derivative of cot inverse is the negative of the derivative of tan inverse. Let us go through the formula of the derivative of cot inverse x in the next section.

Derivative of Cot Inverse x Formula

The formula for the differentiation of cot inverse is given by, d(cot^-1x)/dx = -1/(1 + x²). We can easily memorize the formula of the derivative of arccot using the fact that it is negative of the derivative of arctan. Hence, we can write it as d(cot^-1x)/dx = - d(tan^-1x)/dx = -1/(1 + x²). The image below shows the formula for the derivative of cot inverse x.

Derivative of Cot Inverse x Proof

Now that we know that the derivative of cot inverse x is equal to d(cot^-1x)/dx = -1/(1 + x²), we will prove it using the method of implicit differentiation. Assume y = cot^-1x, then taking cot on both sides of the equation, we have cot y = x. Now, differentiating both sides of the equation cot y = x with respect to x, we have

cot y = x

⇒ -cosec²y × dy/dx = dx/dx

⇒ dy/dx = 1 × -1/cosec²y

⇒ dy/dx = -1/cosec²y

Now, using the trigonometric identity 1 + cot²A = cosec²A, we have

dy/dx = -1/cosec²y

⇒ dy/dx = -1/(1 + cot²y)

= -1/(1 + x²) --- [Because cot y = x]

⇒ d(cot^-1x)/dx = -1/(1 + x²)

Therefore, we have proved that the derivative of cot inverse x is equal to -1/(1 + x²).

Differentiation of Cot Inverse x By First Principle

In this section, we will derive the formula for the derivative of cot inverse using the first principle of derivatives. We will use certain differentiation formulas and inverse trigonometric identities to prove the result such as:

cot^-1x + tan^-1x = π/2 ⇒ cot^-1x = π/2 - tan^-1x
f'(x) = lim_h→0 [f(x+h) - f(x)]/h
d(tan^-1)/dx = 1/(1 + x²)

Using the above formulas, we have

d(cot^-1x)/dx = lim_h→0 [cot^-1(x+h) - cot^-1x]/h

= lim_h→0 [π/2 - tan^-1(x+h) - (π/2 - tan^-1x)]/h

= lim_h→0 [π/2 - tan^-1(x+h) - π/2 + tan^-1x)]/h

= lim_h→0 [- tan^-1(x+h) + tan^-1x)]/h

= - lim_h→0 [tan^-1(x+h) - tan^-1x)]/h

= - d(tan^-1x)/dx ---- [Using the definition of derivatives using limits]

= - 1/(1 + x²) --- [Because the derivative of arctan is 1/(1 + x²)]

Hence, the derivative of cot inverse x is equal to - 1/(1 + x²) using the first principle of derivatives.

Important Notes on Derivative of Cot Inverse

The derivative of cot inverse x is equal to d(cot^-1x)/dx = -1/(1 + x²).
We can derive the cot inverse derivative using the implicit differentiation and first principle of derivatives.

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Derivative of Cot Inverse Examples

Example 1: Evaluate the derivative of cot inverse x square.

Solution: To find the derivative of cot inverse x square, that is, cot^-1(x²), we will use the method of implicit differentiation. Assume y = cot^-1(x²) which implies cot y = x². Now, differentiating cot y = x²with respect to x, we have

d(cot y)/dx = d(x²)/dx

⇒ -cosec²y × dy/dx = 2x

⇒ dy/dx = -2x/cosec²y

= -2x / (1 + cot²y) --- [Using trignometric identity 1 + cot²A = cosec²A]

= -2x / (1 + (x²)²)

= -2x / (1 + x⁴)

We can find this derivative using the derivative of cot inverse x as well.

Answer: Therefore, the derivative of cot inverse x square is given by d(cot^-1(x²))/dx = -2x / (1 + x⁴).
Example 2: Find the derivative of cot inverse [(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))].

Solution: To find the derivative of cot inverse [(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))], we will first rationalize the denominator to simplify the expression and then find its derivative. Consider the expression [(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))], and multiply and divide the expression by (√(1+sinx)+√(1-sinx)).

[(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))] = [(√(1+sinx)+√(1-sinx)) / (√(1+sinx)-√(1-sinx))] × (√(1+sinx)+√(1-sinx)) / (√(1+sinx)+√(1-sinx))

= [(√(1+sinx)+√(1-sinx)) × (√(1+sinx)+√(1-sinx)) ] / [ (√(1+sinx)-√(1-sinx)) × (√(1+sinx)+√(1-sinx)) ]

= (√(1+sinx)+√(1-sinx))² / [(1+sinx) - (1-sinx)] --- [Using a² - b² = (a+b) (a-b)]

= [ (1+sinx) + (1-sinx) + 2 √((1+sinx)(1-sinx)) ] / 2 sinx

= (2 + 2 √(1 - sin²x)) / 2 sinx

= 2 (1 + √cos²x) / 2 sinx --- [Using sin²x + cos²x = 1 ⇒ 1 - sin²x = cos²x]

= (1 + cosx) / sinx

= 2 cos²(x/2) / 2 sin(x/2) cos(x/2) --- [Using cos2A and sin2A formulas: cos2x = 2cos²x - 1 ⇒ cosx = 2cos²(x/2) - 1 ⇒ 1 + cosx = 2cos²(x/2), sin2x = 2 sin x cos x ⇒ sinx = 2 sin(x/2) cos(x/2) ]

= cos(x/2) / sin(x/2)

= cot(x/2)

Therefore, we can write cot^-1[(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))] = cot^-1(cot (x/2)) = x/2

Hence, the derivative of cot^-1[(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))] is given by, d(x/2)/dx = 1/2

Answer: The derivative of cot inverse [(√(1+sinx)+√(1-sinx))/(√(1+sinx)-√(1-sinx))] is 1/2.
Example 3: Find the derivative of cot inverse root x using the formula of the derivative of cot inverse x.

Solution: We know that the derivative of cot inverse x is equal to -1/(1 + x²). Using the chain rule method, we have

d(cot^-1√x)/dx = -1/(1 + √x²) × d(√x)/dx

= -1/(1 + x) × (-1/2√x)

= 1/2 √x (1+x)

Answer: The derivative of cot inverse root x is 1/2 √x (1+x).

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Derivative of Cot Inverse Questions

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FAQs on Derivative of Cot Inverse

What is the Derivative of Cot Inverse in Calculus?

What is the Formula for the Derivative of Cot Inverse x?

The formula for the differentiation of cot inverse is given by, d(cot^-1x)/dx = -1/(1 + x²). We can also write it as d(cot^-1x)/dx = - d(tan^-1x)/dx = -1/(1 + x²)

How to Find the Differentiation of Cot Inverse?

We can evaluate the differentiation of cot inverse using the first principle of derivatives and the method of implicit differentiation.

How is Derivative of Cot Inverse Related to the Derivative of Tan Inverse?

The derivative of cot inverse is the negative of the derivative of tan inverse. Therefore, we can write it as d(arccot)/dx = -d(arctan)/dx = -1/(1 + x²).

What is the Derivative of Tan Inverse x with Respect to Cot Inverse x?

On evaluation, we have d(tan^-1x)/d(cot^-1x) = -1, that is, the derivative of tan inverse x with respect to cot inverse x is equal to -1.

What is the Derivative of Cot Inverse x Square?

The derivative of cot inverse x square is given by d(cot^-1(x²))/dx = -2x / (1 + x⁴). We can calculate this derivative using the derivative of cot inverse x and chain rule method of differentiation.

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