Introduction to the derivative of arctan
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 arctan derivative formula by applying the first principle of differentiation. In this article, you will learn what the derivative of arctan x is and how to calculate the derivative arctan by using different approaches.
What is the derivative of arctan?
The derivative arctan with respect to the variable x' can be expressed as 1 / (1 + x^2). This is commonly denoted as d / dx (arctan x) or d / dx (tan-1x). The arctan function represents the inverse tangent and is the angle whose tangent is equal to x. It is important to note that the art represents the slope of the tangent line to the function at a specific point. Mathematically, it is expressed as;
$tan^{-1} x \;=\; y \;\;or\;\; x \;=\;tan\;y $
Differentiation of arctan formula
The derivative of arc tangent, or the rate of change of arctan x with respect to x, can be calculated using the formula:
$\frac{d}{dx}(arctan x) \;=\; \frac{d}{dx}(tan^{-1} x) \;=\; \frac{1}{1+x^2} $
This formula is essential for calculating the rate of change of arctan x with respect to x. By using this formula, you can easily find the arctan x derivative without any difficulty.
How do you prove the arctan differentiation?
There are different ways or methods to derive derivatives of tan inverse x. These methods allow you to find the rate of change or slope of the tan inverse function at any point, which is useful in many applications. We can prove the derivative of arc tangent by using following derivative rules;
First Principle
Chain Rule
Arctan derivative by first principle
The derivative first principle tells that the differentiation of arctan is equal to the 1/1+x^2. The derivative of a function by first principle refers to finding the slope of a curve by using algebra. It is also known as the delta method. Mathematically, the first principle of derivative formula is represented as:,
$f(x) \;=\; \frac{f(x+h)-f(x)}{h}$
Derivative arctan proof by first principle
To prove the derivative of tan x by using first principle, replace f(x) by tan^-1(x).
$f (x) \;=\; \lim_{h \to 0} \frac{arctan (x+h) - arctan x}{h} $
Now, by using trigonometric formula, tan-1 x-tan-1 y =tan-1 x-y 1+xy , so,
$f (x) \;=\; \lim_{h \to 0}[(\frac{1}{h})\;arctan (\frac{x+h-x}{1+x(x+h)}] $
Simplifying,
$f (x) \;=\; \lim_{h \to 0}[(\frac{1}{h})\;arctan (\frac{h}{1+x^2+xh)}] $
By Taylor's Series, using the expansion of arctan x,
$f (x) \;=\; \lim_{h \to 0}\frac{1}{h}[(\frac{h}{1+x^2+xh}) - \frac{1}{3}(\frac{h}{1+x^2+xh})^3 + \frac{1}{5}(\frac{h}{1+x^2+xh})^5 + ...]$
Simplifying,
$f (x) \;=\; \lim_{h \to 0}[(\frac{1}{1+x^2+xh}) - \frac{1}{3} \frac{h^2}{(1+x^2+xh)^3} + \frac{1}{5}\frac{h^4}{(1+x^2+xh)^5} + ...]$
As h approaches zero, all higher order power will be zero.
$f(x)\;=\;[(\frac{1}{1+x^2+0} + 0)]$
Therefore,
$f(x) \;=\; \frac{1}{1+x^2}$
Our derivative by definition calculator also uses the same method to find the derivative of any function. You can use it online as it is available free to use.
Derivative of arctanx by chain rule
The calculation of the arctan derivative formula involves the use of the chain rule because the tangent function is a combination of two functions. When dealing with a function that can be expressed as y=f(g(x)), the derivative of the chain rule is defined as:
$\frac{dy}{dx} \;=\; \frac{dy}{du} \times \frac{du}{dx}$
Proof of derivative of arctanx by chain rule
To prove derivative of inverse tangent x by using chain rule, consider that,
$tan^{-1} x \;=\; y$
Or we can write it as;
$x \;=\; tan y $
Applying derivative on both sides, we have,
$\frac{d}{dx}(x) \;=\; \frac{d}{dx}(tan y) $
$ 1 \;=\; sec^2y. \frac{dy}{dx} $
By chain rule,
$\frac{dy}{dx}\;=\;\frac{1}{sec^2 y} $
Since,
$sec^2y \;=\; 1 \;+\; tan^2y $
Then,
$\frac{dy}{dx} \;=\; \frac{1}{1+tan^2y} $
Now using the value of x, we get,
$\frac{dy}{dx}\;=\;\frac{1}{1+x^2}$
Use our chain rule calculator to find the derivaitve of tan inverse easily. It will provide you an easy and fast solution for better understanding.
How to find the arctan differentiation with a calculator?
The easiest way to calculate the arctan derivative is by using an online tool. You can use our derivative calculator for this. Here are the steps to calculate the derivative tan using this tool:
Write the function as arctan x or tan-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 arctan(x).
Now, select the variable by which you want to differentiate arctan x. Here you have to choose x'.
Select how many times you want to differentiate arctan (inverse tangent x). In this step, you can choose 2 for second and 3 to find the third derivative calculator.
Click on the calculate button.
After completing these steps, you will receive the derivative of arctangent within seconds. Using online tools can make it much easier and faster to calculate derivatives, especially for complex functions.
Frequently Asked Questions
What is arctan in calculus?
It is the inverse of the trigonometric function tangent. The tangent is the slope of a curve or a line. It is the ratio of the side opposite an angle divided by the adjacent side to that angle. It is written as;
$tan^{-1} x \;=\; y \;\;or\;\; x \;=\;tan\;y $
Is arctan differentiable?
Yes, it is differentiable in its domain. Because, in the domain of arctan, its derivative exists at every point. It is used to determine the measure of angle between the perpendicular length to the base length.
What is arctan formula?
The arctan is the inverse of the tangent function which is the ratio of perpendicular to the base of a triangle. Therefore, the formula of inverse tan is;
$\theta=\tan^{-1}\frac{Base}{\hypotenuse}
What is the derivative of arctan y x?
The derivative of arctanx is calculated the same as the derivative of tan. The formula of derivative of arctangent x is equal to,
$\frac{d}{dx}(arctan x) \;=\; \frac{d}{dx}(tan^{-1} x) \;=\; \frac{1}{1+x^2} $
