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

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.

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.