Introduction to the derivative of cot inverse x

Derivatives have a wide range of applications in almost every field of engineering and science. All derivatives of trigonometric functions can be found by following the derivative of sin x and cos x. Or, we can directly find the derivative of arccot x by applying the first principle of differentiation. In this article, you will learn what the cot inverse derivative is and how to calculate the cot inverse x differentiation by using different approaches. 

What is the differentiation of cot inverse x?

The derivative of arccot x, also known as cot^-1 x, with respect to x is -1/(1+x^2). This formula is essential for computing rates of change in various mathematical applications.

The inverse tangent function represents the angle whose tangent is x, and it is the inverse slope of a line at the point of change in the function. This concept is expressed using the equation;

$\cot^{-1}x=y$ or $x=\cot y$.

Understanding the derivative of arccot x and its applications is crucial in many fields, including engineering, physics, and finance. For a comprehensive explanation of the formula and its usage, read on.

Derivative of arccot x formula

The derivative of the inverse cotangent of x, also known as arccot x, is given by the formula:

$\frac{d}{dx}(arccot x) =\frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1+x^2}$

This formula is important in calculus and related fields for calculating rates of change and slopes of curves. The negative sign in the formula indicates that the slope of the inverse cotangent function decreases as x increases. By understanding and applying this formula, you can solve a wide range of problems in mathematics, physics, and engineering.

How do you prove the derivative of arccotx?

There are multiple ways to derive derivatives of cot x. Two commonly used method are;

1. First Principle
2. Chain Rule

Each method provides a different way to compute the differentiation of cot inverse x. By using these methods, we can mathematically prove the formula for finding the cot inverse derivative.

Cot inverse x derivative by first principle

According to the first principle of derivative, the ln cot^-1x derivative is equal to -1/1+x^2. The derivative of a function by the first principle refers to finding a general expression for the slope of a curve by using algebra. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to,

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

This formula allows us to determine the rate of change of a function at a specific point by using the limit definition of the derivative. You can also use our derivative by definition calculator as it also follows the above formula. 

Proof of derivative of cot-1x by first principle

To prove the cot inverse x differentiation by using first principle, we start by replacing f(x) by cot inverse x. 

$f'(x)=\lim_{h\to 0}\frac{\cot^{-1}(x+h)-\cot^{-1}x}{h}$

Now, by using trigonometric formula,

$\cot^{-1}x+\tan^{-1}x=\frac{\pi}{2}$

So,

$f'(x)=\lim_{h\to 0}\frac{1}{h}\left[\frac{\pi}{2}-\tan^{-1}(x+h)-(\frac{\pi}{2}-\tan^{-1}x)\right]$

Simplifying,

$f'(x)=\lim_{h\to 0}\frac{1}{h}\left[\frac{\pi}{2}-\tan^{-1}(x+h)-\frac{\pi}{2}+\tan^{-1}x\right]$

Simplifying, 

$f'(x)=\lim_{h\to 0}\frac{1}{h}\left[-\tan^{-1}(x+h)+\tan^{-1}x\right]$

$f'(x)=\lim_{h\to 0}\frac{-1}{h}\left[\tan^{-1}(x+h)-\tan^{-1}x\right]$

As h approaches zero,

$\lim_{h\to 0}\frac{\tan^{-1}(x+h)-\tan^{-1}x}{h}=\frac{1}{1+x^2}$

Which is the derivative of tan inverse x. Therefore, 

$f'(x)=-\frac{1}{1+x^2}$

Derivative of arccotx by chain rule

To calculate the cot inverse x differentiation, we can use the chain rule of derivatives since the cot function can be expressed as a combination of two functions. The chain rule of derivatives states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. The chain rule of derivatives is defined as;

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

Proof of derivative of cot inverse x by chain rule 

To prove cot inverse x derivative by using chain rule, we start by assuming that, 

$\cot^{-1}x =y$

Or we can write it as;

$x=\cot y$

Applying derivative on both sides, we have, 

$\frac{d}{dx}(x) =\frac{d}{dx}(\cot y)$

$\implies 1 =-\csc^2y\frac{dy}{dx}$

By the chain rule calculator, 

$\frac{dy}{dx}=-\frac{1}{\csc^2y}$

Since,

$\cot^2y+1=\csc^2y$

Then, 

$\frac{dy}{dx}=-\frac{1}{1+\cot^2y}$

Now using the value of x, we get,

$\frac{dy}{dx}=-\frac{1}{1+x^2}$

Hence, you can derive the derivative of inverse cot by using the chain rule calculator

How to find the derivative of arccot x with a calculator?

The easiest way to calculate the derivative of arccot x is by using an online tool. You can use our derivative calculator for this. Here, we provide you a step-by-step way to calculate derivatives by using this tool.

1. Write the function as arccot x or cot-1x in the “enter function” box. In this step, you need to provide input value as a function as you have to calculate the derivative of cot inverse x.
2. Now, select the variable by which you want to differentiate arccot x. Here you have to choose ‘x’.
3. Select how many times you want to differentiate inverse cot inverse x. In this step, you can choose 2 for the second derivative and 3 to find the third derivative.
4. Click on the calculate button. After this step, you will get the differentiation of cot inverse x within a few seconds.

Frequently asked questions

What is arccot in calculus?

It is the inverse of the trigonometric function cotangent. Since the tangent is the slope of a curve or a line. The cotangent is the reciprocal of the tangent function. It is written as;

$\cot^{-1}x =y\quad \text{or}\quad x=\cot y=\frac{1}{\tan y}$

Is arccot differentiable?

Yes, it is differentiable in its domain. Because, in the domain of arc-cot, its derivative exists at every point. It is used to determine the measure of angle between the perpendicular length to the base length.