Derivative of the arc tangent function is denoted as tan^-1(x) or arctan(x). It is equal to  1/(1+x²). Derivative of arc tangent function is found by determining the rate of change of arc tan function with respect to the independent variable. The technique for finding derivatives of trigonometric functions is referred to as trigonometric differentiation.

Derivative of Arctan

In this article, we will learn about the derivative of arc tan x and its formula including the proof of the formula. Other than that, we have also provided some solved examples for better understanding.

Derivative of Arctan x

Derivative of arc tangent function or arctan(x) is 1/(1+x²). The arctan x represents the angle whose tangent is x. In other words, if y = arctan(x), then tan(y) = x.

The derivative of a function can be found using the chain rule. If you have a composite function like arctan(x), you differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function.

Derivative of Arctan x Formula

The formula for the derivative of inverse of tan x is given by:

d/dx(arctan(x)) = 1/(1+x²)

Also Check:

• Arctan – Formula, Graph, Identities, Domain, Range & FAQs
• Calculus in Maths
• Inverse Trigonometric Function

Proof of Derivative of Arctan x

The derivative of inverse of tan x can be proved using the following ways:

• Using Chain Rule
• Using Implicit Differentiation Method
• Using First Principles of Derivatives

Derivative of Arctan x by Chain Rule

To prove derivative of Arctan x by chain rule, we will use basic trigonometric and inverse trigonometric formula:

• sec²y = 1 + tan²y
• tan (arctan x) = x

Here is the proof of derivative of arctan x:

Let us assume, y = arctan(x)

Taking tan on both sides we get:

tan y = tan(arctan x)

tan y = x [as tan (arctan x) = x]

Now differentiate both sides with respect to x

d/dx (tan y) = d/dx(x)

d/dx(tan y) = 1 [as d/dx(x) = 1]

Applying the chain rule to differentiate tan y with respect to x we get

d/dx(tan y) = sec²y · dy/dx = 1

dy/dx = 1/sec²y

dy/dx = 1/ 1 + tan²y [as sec²y = 1 + tan²y]

Now, we know tan y = x, substituting the value in the above equation we get

dy/dx = 1/ 1 + x²

Derivative of Arctan x by Implicit Differentiation Method

The derivative of arctan x can be proved using the implicit differentiation method. We will use basic trigonometric formulas which are listed below:

• sec²x = ( 1 + tan²x )

• If y = arctan x ⇒ x = tan y and x² = tan²y

Let’s start the proof for the derivative of arctan x , assume f(x) = y = arctan x

By Implicit Differentiation Method

f(x) = y = arctan x

⇒ x = tan y

Taking derivative on both sides with respect to “x”

⇒ d/dx[x] = d/dx[tan y]

⇒ 1 = d/dx[tan y]

Multiplying and dividing the right-hand side by “dy”

⇒ 1 = d/dx[tan y] × dy/dy

⇒ 1 = d/dy[tan y] × dy/dx

⇒ 1 = sec²y × dy/dx

⇒ dx/dy = ( 1+tan²y) [As sec²x = ( 1 + tan²x )]

⇒ dy/dx = 1/( 1+tan²y )

⇒ dy/dx = 1/( 1 + x²) = f'(x)

Therefore f'(x) = 1/ ( 1+x² )

Derivative of Arctan x by First Principle

To prove derivative of arctan x using First Principle of Derivative, we will use basic limits and trigonometric formulas which are listed below:

• lim_h→0 arctan x/x = 1

• arctan x – arctan y = arctan [(x – y)/(1 + xy)]

Let’s start the proof for the derivative of arctan x

we have arctan(x) = y

Apply the definition of derivative we get

[Tex] \frac{d arctan x}{dx} =\displaystyle \lim_{h \to 0} \frac{arctan (x + h)- arctan x}{h}[/Tex]

[Tex] \frac{d arctan x}{dx} =\displaystyle \lim_{h \to 0} \frac{arctan( \frac {x + h – x}{1 + (x + h)x})}{h}[/Tex]

[Tex] \frac{d arctan x}{dx} =\displaystyle \lim_{h \to 0} \frac{arctan( \frac { h}{1 + (x + h)x})}{h\times \frac{1 + (x+h)x}{1 + (x + h)x}}[/Tex]

[Tex]\frac{d arctan x}{dx} =\displaystyle \lim_{h \to 0} \frac{arctan( \frac {h}{1 + (x + h)x})}{(1+(x+h)x)\times \frac{h}{1 + (x + h)x}}[/Tex]

[Tex]\frac{d arctan x}{dx} =\displaystyle \lim_{h \to 0} \frac{1}{(1 +(x+h)x)} \times \displaystyle \lim_{ h\to 0}\frac{arctan\frac{h}{1+(x+h)x}}{\frac{h}{1+(x+h)x}}[/Tex]

[Tex]\frac{d arctan x}{dx} =\displaystyle \lim_{h \to 0} \frac{1}{(1 +x^2+hx)} \times 1[/Tex]

[Tex]\frac{d arctan x}{dx} = \frac{1}{(1 +x^2)}[/Tex]

Also Check

• Derivative of Inverse Trigonometric Functions
• Differentiation Formulas
• Inverse Trigonometric Identities

Examples on Derivative of Arctan x

Example 1: Find the derivative of the function f(x) = arctan(3x).

Solution:

We will use the chain rule, which states that if g(x) is differentiable at x and f(x) = arctan (g(x)), then the derivative f'(x) is given by:

f'(x) = g'(X)/(1+[g(x)]²)

In this case, g(x) = 3x, so g'(X) = 3. Applying the chain rule formula:

f'(x) = 3/(1+(3x)²)

f'(x) = 3/(1+9x²)

Example 2: Find the derivative of the function h(x) = tan^-1(x/2)

Solution:

We will use the chain rule, according which f(x) = tan^-1(g(x)), then the derivative f'(x) is given by:

f'(x) = g'(X)/(1+[g(x)]²)

In this case, g(x) = x/2, so g'(X) = 1/2. Applying the chain rule formula:

f'(x) = (1/2)/(1+(x/2)²)

f'(x) = (1/2)/(1+x²/4)

Simplifying we get,

f'(x) = 2/(4+x²)

Example 3: Find the derivative of f(x) = arctan (2x²)

Solution:

We will use the chain rule, which states that if g(x) is differentiable at x and f(x) = arctan (g(x)), then the derivative f'(x) is given by:

f'(x) = g'(X)/(1+[g(x)]²)

In this case, g(x) = 2x², so g'(X) = 4x.

Applying the chain rule formula:

f'(x) = 4x/(1+(2x²)²)

f'(x) = 4x/(1+4x⁴)

f'(x) = d/dx(arctan (2x²)) = 4x/(1+4x⁴)

Practice Questions on Derivative of Arctan x

Q.1: Find the derivative of the function f(x) = x²arcan (2x)

Q.2: Find the derivative of the function k(x) = arctan (x³+2x)

Q.3: Find the derivative of the function p(x) = x arctan(x²+1)

Q.4: Find the derivative of the function f(x) = arctan (x)/1+x

Q.5: Find the derivative of the function r(x) = arctan (4x)

Read More,

• Derivative in Math
• Derivative of tan inverse x
• Arctan

Summary

The derivative of the arctangent function, denoted as[Tex] \frac{d}{dx} (\arctan(x))[/Tex], is given by [Tex]\frac{1}{1+x^2}[/Tex]​. This result can be derived using implicit differentiation and trigonometric identities. Starting with [Tex]y = \arctan(x)[/Tex], we can take the tangent of both sides to get [Tex]x = \tan(y)[/Tex]. Differentiating both sides with respect to x yields [Tex]\sec^2(y) \frac{dy}{dx}[/Tex]​. Using the identity [Tex]\sec^2(y) = 1 + \tan^2(y)[/Tex] and substituting [Tex]\tan(y) = x[/Tex] back in, we find [Tex]\frac{dy}{dx}1=(1+x^2)dxdy[/Tex]​, leading to [Tex]\frac{dy}{dx} = \frac{1}{1 + x^2}dxdy[/Tex]​. Thus, the derivative of [Tex]\arctan(x)[/Tex] is [Tex]\frac{1}{1 + x^2}[/Tex]​.

Derivative of Arctan x – FAQs

What is Derivative in Math?

In mathematics, the derivatives the measures how a function changes as its input (independent variable) changes.The derivative of a function f(x) is denoted as f'(x) or (d /dx)[f(x)].

What is Derivative of tan^-1(x)?

Derivative of the tan^-1(x) with respect to x is 1/1+x²

What is Inverse of tan x?

Arctan is the inverse of tan function and it is one of the inverse trigonometric functions. It is also known as the arctan function.

What is Chain Rule in Arctan (x)?

Chain rule is a differentiation rule. For arctan (u), the chain rule states that if f(x) = arctan(u), then f'(x) = (1/1+u²)× du/dx. Applying this to arctan(x), where u=x, gives 1/1+x²

What is Derivative of f(x) = x tan^-1(x)?

Derivative of f(x) = xtan^-1(x) can be found using the product rule. The result is tan^-1(x) + {x/(1 + x²)}.

What is Anti Derivative of Arctan x?

Antiderivative of arctan x is given by ∫tan^-1x dx = x tan^-1x – ½ ln |1+x²| + C.

What is Derivative?

Derivative of function is defined as the rate of change of the function with respect to an independent variable.