Derivative of Tan Inverse x

Derivative of tan inverse x is 1/(1+x²). Derivative of tan inverse x refers to the process of finding the change in the inverse tangent function to the independent variable. The specific process of finding the derivative for inverse trigonometric functions is referred to as inverse trigonometric differentiation, and the derivative of tan^-1x is one of the key results in inverse trigonometric differentiation.

In this article, we will learn about the derivative of tan inverse x and its formula including the proof of the formula using the first principle of derivatives, quotient rule, and chain rule as well.

What is Derivative in Math?

The derivative of a function is the rate of change of the function to any independent variable. The derivative of a function f(x) is denoted as f'(x) or (d /dx)[f(x)]. The differentiation of an inverse trigonometric function is called a derivative of the inverse trigonometric function or inverse trig derivatives.

Read in Detail:

• What are Derivatives?
• Calculus in Maths

What is the Derivative of tan^-1x?

Among the inverse trig derivatives, the derivative of tan^-1x is one of the derivatives. The derivative of tan^-1x is 1/(1+x²). The derivative of tan^-1x is the rate of change to angle, i.e. x. The resultant of the derivative of tan^-1x is 1/(1+x²).

Derivative of tan^-1x Formula

The formula for the derivative of tan^-1x is given by:

(d/dx) [tan^-1x] = 1/ ( 1+x² )

or

(tan^-1x)’ = 1/ ( 1+x² )

Proof of Derivative of TanInverse x

The derivative of tan^-1x can be proved using the following ways:

• By using the First Principle of Derivative

• By using Implicit Differentiation Method

• By using Cot Inverse Formula

Derivative of tan^-1x by First Principle of Derivative

To prove derivative of tan^-1x using First Principle of Derivative, we will use basic limits and trigonometric formulas which are listed below:

• tan^-1a - tan^-1b = tan^-1( ( a-b ) / ( 1+ab ))

• lim _h→0 tan^-1h/h=1

Let’s start the proof for the derivative of tan^-1x , assume f(x)=tan^-1x ⇒

By First Principle of Derivative

f'(x) = d/dx[tan^-1x] = lim_h→0[tan^-1(x+h) - tan^-1(x)]/h

⇒ lim_h→0 tan^-1( ( x+h-x ) / ( 1+x.( x+h ) ) )/h {By 1}

⇒ lim_h→0 tan^-1( h/ ( 1+x.( x+h ) ) )/h

Divide and multiply the above expression with 1/ (1 + x.( x+h))

⇒ lim_h→0 [tan^-1( h/ (1 + x.(x+h)))/h ] × [(1/ ( 1 + x.(x+h)))/(1/ ( 1 + x.(x+h)))]

⇒ lim _h→0 1 × (1/1+x(x+h)) {By 2}

⇒1 × (1/1+x(x+0))

⇒ 1/ (1 + x²)

Therefore f'(x) = 1/ ( 1+x² )

Derivative of tan^-1 x by Implicit Differentiation Method

The derivative of tan^-1 x can be proved using the implicit differentiation method. In this method, if we are given an implicit function, then we take the derivative on both sides of the equation with respect to the independent variable. We will use basic trigonometric formulas which are listed below:

• sec²x = ( 1 + tan²x )

• If y = tan^-1x ⇒ x = tan y and x² = tan²y

Let’s start the proof for the derivative of tan^-1x , assume f(x) = y = tan^-1x

By Implicit Differentiation Method

f(x) = y = tan^-1x

⇒ 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] x dy/dy

⇒ 1 = d/dy[tan y] x dy/dx

⇒ 1 = sec²ydy/dx

⇒ dx/dy = ( 1+tan²y) { By 1 }

⇒ dy/dx = 1/( 1+tan²y )

⇒ dy/dx = 1/( 1 + x²) = f'(x) { By 2 }

Therefore f'(x) = 1/ ( 1+x² )

Derivative of tan^-1 x by Cot^-1 x Formula

The derivative of tan^-1 x can be proved by using another trigonometric inverse function of cot^-1x. We will differentiate tan^-1 x with respect to cot^-1x using basic trigonometric formulas which are listed below:

• cot²x + 1 = cosec²x

• sec²x = ( 1 + tan²x )

Let’s start the proof for the derivative of tan^-1x , assume f(x)=y=tan^-1x

By Cot^-1 x Formula,

f(x) = y = tan^-1x

⇒ x = tan y

Taking derivative on both sides with respect to “x"

⇒dx/dx = d/dx[tan y]

⇒ 1 = d/dx[ tan y ]

Multiplying and dividing the right-hand side “dy"

⇒ 1 = d/dx[ tan y ] × dy/dy

⇒ 1 = sec²y x dy/dx

⇒ dy/dx = 1/sec²x

⇒ dy/dx = 1/(1+x²) - ( 1 ) { By 2 }

Let g = cot^-1x , i.e. x = cot g

Now differentiating above function with respect to “x"

⇒ dx/dx = d/dx[cot g]

⇒ 1 = d/dx[cot g]

Multiplying and dividing right hand side by “dg"

⇒ 1 = d/dx[cot g] × dg/dg

⇒ 1 = -cosec²g × dg/dx

⇒ dx/dg = - cosec²g

⇒ dx/dg = - (1+x²) - (2) {By 1 }

We need to find out the derivative of tan^-1x with respect to cot^-1x , which is dy/dg

dy/dg=(dy/dx) × (dx/dg)

⇒ [1/(1+x²)] × [- (1+x²)] {From eqn. 1 and 2 }

⇒ -1 = dy/dg

We know that d tan^-1x / d cot^-1 x = -1 .Hence indirectly we can say that derivative of tan^-1x is 1/(1+x²).

Therefore f'(x) = 1/ ( 1+x² )

Learn More,

• Inverse Trigonometric Function
• Differentiation Formulas
• Inverse Trigonometric Identities

Examples on Derivative of TanInverse x

Example 1: Find the derivative of tan^-1(x²).

Solution:

d/dx[tan^-1(x²)] = 1/(1+(x²)²) × 2x = 2x/(1+x⁴)

Example 2: Find the derivative of tan^-1(x) at x = 1.

Solution:

d/dx[tan^-1x] = 1/(1+x²)

at x=1 , d/dx[tan^-1x] = 1/(1+1²) = 1/2

Example 3: Find the derivative of tan^-1(1/x)

Solution:

d/dx[tan^-1(1/x)] = 1/( 1 + (1/x)²) x -1/x²

⇒[-1/x²] × [x²/( 1+x² ) ]

⇒-1/( 1+x² ) = d/dx[tan^-1(1/x)]

Example 4: Find the derivative of tan^-1(x³)

Solution:

d/dx[tan^-1(x³)] = 1/(1+(x³)²) × (3x) = 3x/(1+x⁶)

Practice Questions on Derivative of tan^-1x

Q1. Find the derivative of tan^-1 7x

Q2. Find the derivative of x².tan^-1x

Q3. Evaluate: (d/dx) [tan^-1x/(x² + 2)]

Q4. Evaluate the derivative of: cot^-1x. tan^-1 x

Q5. Find: (tan^-1x)sin x

Q6. Find the derivative of sin^-1(2x)

Q7. Evaluate d/dx [ x / sqrt(1 - x²) ]

Q8. Find the derivative of e^tan-1(x)

Q9. Compute d/dx [ ln(sec(x) + tan(x)) ]

Q10. Find the derivative of cos^(-1(x) / (x² + 1)

Conclusion

The derivative of Tan^-1x , a fundamental result in inverse trigonometric differentiation. This result is derived using various methods, including the first principle of derivatives, implicit differentiation, and relationships with other inverse trigonometric functions. Understanding this derivative is crucial for solving problems involving rates of change and integration in calculus. The derivative captures the rate of change of the inverse tangent function with respect to (x), reflecting how the function's slope varies with (x).