Derivative of Tan Inverse x

The derivative of tan inverse x is given by (tan^-1x)' = 1/(1 + x²). The differentiation of tan inverse x is the process of finding the derivative of tan inverse x with respect to x. The derivative of tan inverse x can also be interpreted as the rate of change of tan inverse x which is given by 1/(1 + x²).

In this article, we will learn the concept of the derivative of arctan, its proof using implicit differentiation, the first principle of differentiation, and the derivative of tan inverse x with respect to cot inverse x along with some examples for a better understanding.

Differentiation of tan inverse x is the process of evaluating the derivative of tan inverse x with respect to x which is given by 1/(1 + x²). The derivative of tan inverse x can be calculated using different methods such as the first principle of derivatives and using implicit differentiation. Since the derivative of arctan with respect to x which is 1/(1 + x²), the graph of the derivative of arctan is the graph of algebraic function 1/(1 + x²)

Derivative of Tan Inverse x Formula

An easy way to memorize the derivative of tan inverse x is that it is the negative of the derivative of cot inverse x. In other words, we can say the derivative of cot inverse x is negative of the derivative of tan inverse x. The formula for the derivative of tan inverse x is given by,

d(tan^-1x)/dx = 1/(1 + x²)

To prove the derivative of tan inverse x using implicit differentiation, we will use the following trigonometric formulas and identities:

• d(tan x)/dx = sec²x
• sec²x = 1 + tan²x
• tan(tan^-1x) = x

Using the above formulas, we have

Given: f(x) = tan^-1x

Let y = tan^-1x

⇒ tan y = x (Using tan(tan^-1x) = x)

⇒ x = tan y ---- (1)

Differentiate both sides of x = tan y w.r.t. x

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

⇒ 1 = [d(tan y)/dx] × [dy/dy] [Multiplying and dividing by 'dy']

⇒ 1 = [d(tan y)/dy] × [dy/dx]

⇒ 1 = [sec²y] dy/dx (Using d(tan x)/dx = sec²x)

⇒ 1 = [1 + tan²y] dy/dx [Using trigonometric identity, sec²y = 1 + tan²y]

dy/dx = [1] / [1 + tan²y]

Substituting tan y = x (from (1)) into dy/dx = [1] / [1 + tan²y], we have

dy/dx = 1 / (1 + x²)

⇒ d(tan^-1x)/dx = 1 / (1 + x²)

Hence, we have derived the derivative of tan inverse x using implicit differentiation.

Now we will evaluate the derivative of arctan using the first principle of differentiation. To derive it, we will use some differentiation and trigonometric formulas and identities such as:

• \(f'(x)=\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}\)
• \(\lim_{h\rightarrow 0}\dfrac{tan^{-1}x}{x} = 1\)
• tan^-1x - tan^-1y = tan^-1[(x - y)/(1 + xy)]

Using the above formulas, we have

\(\begin{align}\frac{\mathrm{d} \tan^{-1}x}{\mathrm{d} x} &=\lim_{h\rightarrow 0}\dfrac{ \tan^{-1}(x+h)- \tan^{-1}x}{h}\\&=\lim_{h\rightarrow 0}\dfrac{ \tan^{-1}(\frac{x+h-x}{1+(x+h)x})}{h}\\&=\lim_{h\rightarrow 0}\dfrac{ \tan^{-1}(\frac{h}{1+(x+h)x})}{h\times\frac{1+(x+h)x}{1+(x+h)x}}\\&=\lim_{h\rightarrow 0}\dfrac{ \tan^{-1}(\frac{h}{1+(x+h)x})}{(1+(x+h)x)\times\frac{h}{1+(x+h)x}}\\&=\lim_{h\rightarrow 0}\dfrac{1}{1+(x+h)x}\times \lim_{h\rightarrow 0}\dfrac{ \tan^{-1}(\frac{h}{1+(x+h)x})}{\frac{h}{1+(x+h)x}}\\&=\dfrac{1}{1+x^2}\times 1\\&= \dfrac{1}{1+x^2}\end{align}\)

Hence, we have calculated the derivative of tan inverse x using the first principle of derivatives.

Since the derivative of tan inverse x is 1/(1 + x²), we will differentiate tan^-1x with respect to another function, that is, cot^-1x. For this, we will assume cot^-1x to be equal to some variable, say z, and then find the derivative of tan inverse x w.r.t. cot^-1x.

Assume y = tan^-1x ⇒ tan y = x

Differentiating tan y = x w.r.t. x, we get

sec²y (dy/dx) = 1

⇒ dy/dx = 1/sec²y

⇒ dy/dx = 1/(1 + tan²y) (Using trigonometric identity 1 + tan²θ = sec²θ)

⇒ dy/dx = 1/(1 + x²) ---- (1)

Assume z = cot^-1x ⇒ cot z = x

Now, differentiating cot z = x w.r.t. x, we have

-cosec²z (dz/dx) = 1

⇒ dz/dx = -1/cosec²z

⇒ dz/dx = -1/(1 + cot²z) [(Using trigonometric identity 1 + cot²θ = cosec²θ)]

⇒ dz/dx = -1/(1 + x²)

⇒ dx/dz = -(1 + x²) ---- (2)

Now, we need to determine the value of d(tan^-1x)/d(cot^-1x) = dy/dz

dy/dz = dy/dx × dx/dz

= [1/(1 + x²)] × (-(1 + x²))

= -1

Hence, d(tan^-1x)/d(cot^-1x) = -1, that is, derivative of tan inverse x w.r.t. cot inverse x is -1.

Important Notes on Derivative of Acrtan

• The derivative of tan inverse x is that it is the negative of the derivative of cot inverse x.
• The derivative of tan inverse x with respect to x is 1/(1 + x²).
• Anti-derivative of tan inverse x is given by, ∫ tan^-1x dx = x tan^-1x - ½ ln |1+x²| + C

Related Topics on Derivative of Tan Inverse x

• Derivative of sin inverse x
• Derivative of cos 2x
• Derivative of Arccos

What is Derivative of Tan Inverse x in Trigonometry?

The derivative of tan inverse x with respect to x is given by 1/(1 + x²). The derivative of tan inverse x can be calculated using the concept of derivatives and inverse trigonometric functions.

What is the Derivative of Tan Inverse x + Cot Inverse x?

We know that the derivative of tan inverse x is that it is the negative of the derivative of cot inverse x and the derivative of tan inverse x is 1/(1 + x²). Therefore, the derivative of Tan Inverse x + Cot Inverse x is 1/(1 + x²) + [-1/(1 + x²)] = 0.

What is the Derivative of Tan Inverse x With Respect to Cot Inverse x?

The derivative of tan inverse x w.r.t. cot inverse x is -1.

How to Find the Derivative of Arctan?

The derivative of arctan can be calculated using different methods such as implicit differentiation and the first principle of differentiation.

What is the Anti-Derivative of Tan Inverse x?

Anti-derivative of tan inverse x is given by, ∫ tan^-1x dx = x tan^-1x - ½ ln |1+x²| + C