find nth derivative of the funcrtion e^x sin^2 x

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Learn how to find nth derivative of the funcrtion e^x sin^2 x using partial fractions and differentiation techniques. Follow our detailed step-by-step solution to master this calculus problem

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

Solution:- Let y = e^x.\sin^2 x

⇒ y = e^x\dfrac{1 - \cos 2x}{2}

⇒ y = \frac{1}{2}[e^x - e^x\cos 2x]

Differentiate with respect to x

y_1 = \frac{1}{2}[e^x -(e^x\cos 2x - e^x.\sin 2x.2)]

⇒ y_1 = \frac{1}{2}[e^x -e^x(\cos 2x - 2.\sin 2x)]

⇒ y_1 = \frac{1}{2}[e^x - \sqrt{5}e^x(\frac{1}{\sqrt{5}}\cos 2x - \frac{2}{\sqrt{5}}\sin 2x)] .  .  .  (i)

Let \frac{1}{\sqrt{5}} = \cos \theta then  \frac{2}{\sqrt{5}} = \sin \theta

Thus, \tan \theta = \dfrac{2/\sqrt{5}}{1/\sqrt{5}}

⇒ \tan \theta = 2

⇒ \theta = \tan^{-1} 2

From equation (i)

⇒ y_1 = \frac{1}{2}[e^x - \sqrt{5}e^x(\cos \theta\cos 2x -\sin \theta\sin 2x)]

⇒ y_1 = \frac{1}{2}[e^x - \sqrt{5}e^x\cos (2x + \theta)]

Again differentaite with respect to x

y_2 = \frac{1}{2}[e^x- (\sqrt{5})^2e^x\cos(2x + 2\theta)]

Now differentiate with respect to x for y_n

y_n = \frac{1}{2}[e^x- (\sqrt{5})^ne^x\cos(2x + n\theta)]

y_n = \frac{1}{2}[e^x- (\sqrt{5})^ne^x\cos(2x + n\tan^{-1} 2)]

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

Solution:- See full solution

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