The nth derivative of tan inverse (1+x)/(1-x)

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Learn how to find The nth derivative of tan inverse (1+x)/(1-x) using partial fractions and differentiation techniques. Follow our detailed step-by-step solution to master this calculus problem

Q 3:- Find the nth derivative of \tan^{-1}\left[\dfrac{1+x}{1-x}\right]

Solution:- Let y = \tan^{-1}\left[\dfrac{1+x}{1-x}\right]

Let x = \tan \theta

y = \tan^{-1}\left[\dfrac{1+\tan \theta}{1-\tan \theta}\right]

⇒ y = \tan^{-1}\left[\dfrac{\tan \pi/4+\tan \theta}{1-\tan \pi/4\tan \theta}\right]

⇒ y = \tan^{-1}\left[\tan (\pi/4 + \theta)\right]

⇒ y = \pi/4 + \theta

⇒ y = \pi/4 + \tan^{-1} x

Now, differentiate with respect to x

y_1 = 0 + \dfrac{1}{1 + x^2}

⇒ y_1 = \dfrac{1}{ x^2 + 1}

⇒ y_1 = \dfrac{1}{ x^2 - i^2}

⇒ y_1 = \dfrac{1}{ (x+i)(x-i)}

Using partial fraction

\dfrac{1}{ (x+i)(x-i)} = \dfrac{A}{x + i} + \dfrac{B}{x-i} .  .  . (i)

⇒ \dfrac{1}{ (x+i)(x-i)} = \dfrac{A(x-i) + B(x+i)}{(x+i)(x-i)}

⇒ 1 = A(x-i) + B(x+i)

Let x = i

1 = 0 + B(i+i)

⇒ B = 1/2i

Let x = -i

1 = A(-i-i) + 0

⇒ A = -1/2i

Replacing the value of A and B in equation (i)

We get

\dfrac{1}{ (x+i)(x-i)} = \dfrac{-1/2i}{x + i} + \dfrac{1/2i}{x-i}

⇒ y_1 = -\frac{1}{2i}(x+i)^{-1} + \frac{1}{2i}(x-i)^{-1}

Now, differentiate with respect to x

y_2 = -\frac{1}{2i}(-1)(x+i)^{-2} + \frac{1}{2i}(-1)(x-i)^{-2}

Again differentiate with respect to x

y_3 = -\frac{1}{2i}(-1)(-2)(x+i)^{-3} + \frac{1}{2i}(-1)(-2)(x-i)^{-3}

⇒ y_3 = -\frac{1}{2i}(-1)^{2}(2)!(x+i)^{-3} + \frac{1}{2i}(-1)^2(2)!(x-i)^{-3}

Now,

y_n = -\frac{1}{2i}(-1)^{n-1}(n-1)!(x+i)^{-n} + \frac{1}{2i}(-1)^{n-1}(n-1)!(x-i)^{-n}

⇒ y_n = \frac{1}{2i}(-1)^{n-1}(n-1)![(x-i)^{-n} - (x+ i)^{-n}]

Compare (x + i) = r(\cos \theta + i\sin \theta)

r\cos \theta = x and r\sin \theta = 1

r^2 = x^2 + 1^2 and \cot \theta = x

y_n = \frac{1}{2i}(-1)^{n-1}(n-1)![r^{-n}(\cos \theta - i\sin \theta)^{-n} - r^{-n}(\cos \theta + i\sin \theta)^{-n}]

⇒ y_n = \frac{1}{2i}(-1)^{n-1}(n-1)!r^{-n}[(\cos (-n\theta) - i\sin (-n\theta)) -(\cos (-n\theta) + i\sin (-n\theta))]

⇒ y_n = \frac{1}{2i}(-1)^{n-1}(n-1)!r^{-n}[(\cos (n\theta) + i\sin (\theta)) -(\cos (n\theta) - i\sin (n\theta))]

⇒ y_n = \frac{1}{2i}(-1)^{n-1}(n-1)!r^{-n}[\cos (n\theta) + i\sin (\theta) -\cos (n\theta) + i\sin (n\theta))]

⇒ y_n = \frac{1}{2i}(-1)^{n-1}(n-1)!r^{-n}[2 i\sin (\theta)]

⇒ y_n = (-1)^{n-1}(n-1)!r^{-n}\sin n(\theta)

⇒ y_n = \dfrac{(-1)^{n-1}(n-1)!}{r^n}\sin n(\theta)

⇒ y_n = \dfrac{(-1)^{n-1}(n-1)!}{r^n}\sin n(\theta) .  .  . (ii)

Given,  r = \sqrt{x^2 +1}

Since, x = \cot \theta

r = \sqrt{\cot^2 \theta+1}

⇒ r = \sqrt{\operatorname{cosec}^2 \theta}

⇒ r = \operatorname{cosec} \theta

⇒ \dfrac{1}{r} = \sin \theta

⇒ \dfrac{1}{r^n} = \sin^n \theta

Hence from equation (ii)

y_n = (-1)^{n-1}(n-1)! \sin^n \theta\sin n(\theta)

Where, x = \cot \theta

Some other question

Q 1:  If y = \dfrac{x^2}{(x-1)^2(x+2)}, find nth derivative of y.

Solution: See full solution

Q 2: Find the nth derivative of \dfrac{x^2}{(x-a)(x-b)}.

Solution:- See full solution

Q 4:-  If y = sin³ x find nth term of y.

Solution:- See full solution

Q 5:- Find nth derivative of \tan^{-1}(x/a) .

Solution:- See full solution

Q 6:- Find nth derivative of e^x.x .

Solution:- See full solution

Q 7:- Find nth derivative of the function \dfrac{1-x}{1+x}.

Solution:- See full solution

Q 8:- Find nth derivative of \lox x^2

Solution:- Find full solution

Q 9:- Find nth derivative of the funcrtion e^x.\sin^2 x.

Solution:- See full solution

Q 10:- Find the value of nth derivative of y=\cos x\cos 2x \cos 3x.

Solution:- See full solution

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