Find nth derivative of tan inverse (x/a)

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

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

Solution:- Let y = \tan^{-1}(x/a)

Differentiate with respect to x

y_1 = \dfrac{1}{1+(x/a)^2}\dfrac{d}{dx}(x/a)

⇒ y_1 = \dfrac{1}{1+(x/a)^2}\dfrac{d}{dx}(x/a)

⇒ y_1 = \dfrac{a^2}{x^2 + a^2}.(1/a)

⇒ y_1 = \dfrac{a}{x^2 + a^2}

⇒ y_1 = \dfrac{a}{x^2 -i^2 a^2}

⇒ y_1 = \dfrac{a}{(x+ia)(x - ia)}

Using partial fraction

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

⇒ \dfrac{a}{(x+ia)(x - ia)} = \dfrac{A(x - ia) + B(x+ia)}{(x+ia)(x-ia)}

⇒ a = A(x – ia) + B(x+ia)

Let x = ia

a = 0 + B(ia+ia)

⇒ B = a/2ia = 1/2i

Let x = -ia

a = A(-ia-ia) + 0

⇒ A = -a/2ia = -1/2i

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

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

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

Differentiate with respect to x

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

Again differentiate with respect to x

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

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

Now differentiate this equation for y_n

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

⇒ y_n = \dfrac{1}{2i}(-1)^{n-1}(n-1)![(x - ia)^{-n} - (x + ia)^{-n}] .  .  . (ii)

Let (x - ia) = r(\cos \theta - i\sin \theta) and (x + ia) = r(\cos \theta + i\sin \theta)

Now, x = r\cos \theta and a = r\sin \theta

Adding after square

x^2 + y^2 = r^2 and \tan \theta = a/x

Now from equation (ii)

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

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

Since, r^2 = x^2 + a^2 and \tan \theta = a/x

x = a\cot \theta

Now, r^2 = (a\cot \theta)^2 + a^2

⇒ r^2 = a^2(\cot^2\theta + 1)

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

⇒ r = a\operatorname{cosec}\theta

⇒ 1/r^n = \frac{a}{a}\sin^n\theta

⇒ r^{-n} = a^{-n}\sin^n\theta

From equation (iii)

y_n =(-1)^{n-1}(n-1)!a^{-n}\sin^n\theta\sin n\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 3:- Find the nth derivative of \tan^{-1}\left[\dfrac{1+x}{1-x}\right]

Solution:- See full solution

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

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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