The arctan derivative, or the inverse tangent function, is an important notion in calculus and is used as a fundamental technique in differentiation, integration, and trigonometry. Learning the derivation and how to use it assists in breaking down intricate physics, engineering, and computer science problems. Since arctan represents the angle whose tangent is a given value, its derivative is unique and follows a distinct mathematical formula, making it essential for students and professionals working with trigonometric functions.

In this maths article we will understand the concept of derivative of arctan and its proof by first principle and chain rule, with related solved examples. 

Derivative of Arctan(x)

The derivative of the arctan function, often written as [latex]\frac{d}{dx}\left(\arctan(x)\right)[/latex], is the rate of change of the arctan function with respect to the variable [latex]x[/latex]. The derivative of arctanx is equal to [latex]\frac{1}{1+x^2}[/latex]. 

This result allows us to determine the slope or gradient of the arctan function at any point [latex]x[/latex] on its graph.

Arctan is the inverse tangent function, for example, if x and y are any variables then \(y=\tan ^{-1}x\) then \(tan \: y = x\).

Similarly, \(tan (arctan \: x) = arctan (tan \: x)\)

Derivative of Arctan(x) Formula

Formula for derivative of Arctan function is

\(\Large \frac{d}{dx}(\arctan x)=\frac{1}{1+x^2} \text{or}\)   \(\Large \frac{d}{dx}(\tan ^{-1}x)=\frac{1}{1+x^2}\)

Derivative of Arctan(u)

The arctan function can also be applied to a function [latex]u[/latex] of [latex]x[/latex]. When we have [latex]u[/latex] as a function of [latex]x[/latex], the derivative of [latex]\arctan(u)[/latex] with respect to [latex]x[/latex] can be computed using the chain rule. 

The formula for the derivative of [latex]\arctan(u)[/latex] is [latex]\frac{{du}}{{dx}} \cdot \frac{{1}}{{1 + u^2}}[/latex]. 

For example, let's consider [latex]u(x) = 3x^2 + 2x - 1[/latex]. 

The derivative of [latex]\arctan(u(x))[/latex] can be calculated as [latex]\frac{{d}}{{dx}} \arctan(3x^2 + 2x - 1) = \frac{{d}}{{dx}}(3x^2 + 2x - 1) \cdot \frac{{1}}{{1 + (3x^2 + 2x - 1)^2}}[/latex]. 

By differentiating [latex]u(x)[/latex] and substituting it into the formula, we can determine the derivative of [latex]\arctan(u)[/latex] for the given function [latex]u(x)[/latex].

Why is arctan differentiable?

The arctan function, also known as the inverse tangent function, is differentiable because it possesses a smooth and continuous graph. The arctan function is defined for all real numbers and its graph has no jumps.

Derivative of Arctan by Chain Rule

Let’s find the derivative of arctan by using the chain rule,

Step 1: Let’s take the function,  \(y = arctan x\) (as equation 1) where y and x are variables.

Step 2: Taking “tan” function on both sides of equation (1)

• \(tan \: y = tan (arctan \: x)\)
• \(tan (arctan \: x) = x \) (by the law of inverse function)
• Hence, \(tan \: y = x\) (equation 2)

Step 3: Differentiating equation 2 with respect to x,

\(\frac{d}{dx}(\tan \: y)=\frac{d}{dx}(x)\)

Step 4: Now, according to the formula \(\frac{d}{dx}(\tan \: x)=\sec ^2x\) then,

\(\frac{d}{dx}(\tan \: y)=\sec ^2y\)

Step 5: According to the chain rule,

• \(\sec ^2y\times \frac{dy}{dx}=1\)
• \(\frac{dy}{dx}=\frac{1}{\sec ^2y}\)

Step 6: Using one of the trigonometric identities of \(\sec ^2y=1+\tan ^2y\)

We get, \(\frac{1}{1+\tan ^2y}=\frac{dy}{dx}\)

Step 7: Substituting y = arctan x , we finally get

\(\frac{d}{dx}(\arctan \: x)=\frac{1}{1+x^2}\).

Derivative of Arctan by First Principle

To find the derivative of arctan by first principle of derivatives,

Step 1: Let’s consider the function f(x), where x is a variable

Step 2: By the first principle in the given limit, \(f^{\prime}(x)=lim_{h\rightarrow 0}\frac{[f(x+h)-f(x)]}{h}\) where we assume that,

• \(f(x) = arctan \: x\)
• \(f(x+h) = arctan (x+h)\)

Step 3: We get the limit function as,

• \(f^{\prime}(x)=\frac{lim_{h\rightarrow 0}\frac{[(x+h)-(x)]}{1+(x+h)x}}{h}\)
• \(f^{\prime}(x)=lim_{h\rightarrow 0}\frac{1}{h}\frac{h}{1+x^2+hx}-\frac{[\frac{h}{1+x^2+hx}]^3}{3}+\frac{\frac{h}{1+x^2+hx}}{5}…\)
• \(f^{\prime}(x)=lim_{h\rightarrow0}\left[\frac{1}{1+x^2+hx}-\frac{h^2}{3\left(1+x^2+hx\right)^3}+\frac{h^4}{5\left(1+x^2+hx\right)^5}…..\right]\)

Step 4: Applying the evaluation of limits, \(lim_{h\rightarrow0}\), we get

• \(f'(x)=\frac{1} {1+x^2+0}-0+0….\)
• \(f^{\prime}(x)=\frac{1}{1+x^2}\)

Solved Examples of Derivative of Arctan

Example 1: In a right-angled triangle if the base of the triangle is 5 units and the height of the triangle is 7 units, find the base angle.

Solution: In a right-angled triangle

The formula to find a base angle is given by,

\(\theta =\tan ^{-1}\frac{7}{5}\)

\(\theta =\tan ^{-1}(1.4)\)

= arctan (1.4)

= \(54.5^{\circ }\).

Hence the base angle of a right-angled triangle is \(54.5^{\circ }\).

Example 2: Find the derivative of y=arctan (4x).

Solution: Let f(x) = arctan (4x)

We know that,

\(\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}\)

Hence by chain rule,

\(y’=\frac{1}{1+4x^2}.\frac{d}{dx}(4x)\)

\(y’=\frac{1}{1+4x^2}.4(1)\)

\(y’=\frac{4}{1+4x^2}\) is the derivative of function 4x.

Example 3: Find the derivative of \(y=\arctan \frac{1}{x}\)

Solution: \(f(x)=\arctan \frac{1}{x}\)

Hence \(\frac{d}{dx}(\arctan x)=\frac{1}{1+x^2}\)

By the chain rule,

\(f(x)=\frac{1}{1+\left(\frac{1}{x}\right)^2}.\frac{d}{dx}\frac{1}{x}\)

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

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

The derivative of the given function is \(f(x)=\frac{-1}{x^2+1}\)

Mastery of the derivative of arctan is essential in overcoming complex mathematical challenges, particularly in calculus and engineering sciences. Its smooth and well-behaved form lends itself well to practical applications such as signal processing and machine learning. Mastery of the derivative of arctan enables students to enhance their problem-solving ability and apply these principles in advanced mathematics with assurance.