Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

X Sin N X - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads

\[\frac{x}{\sin^n x}\]

Advertisement Remove all ads

Solution

\[\text{ Let } u = x; v = \sin^n x\]
\[\text{ Then }, u' = 1; v' = n \sin^{n - 1} x . \cos x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{x}{\sin^n x} \right) = \frac{\sin^n x . 1 - x n \sin^{n - 1} x . \cos x}{\left( \sin^n x \right)^2}\]
\[ = \frac{\sin^{n - 1} x\left( \sin x - nx . \cos x \right)}{\sin^{2n} x}\]
\[ = \frac{\sin x - nx . \cos x}{\sin^{2n - \left( n - 1 \right)} x}\]
\[ = \frac{\sin x - nx\cos x}{\sin^{n + 1} x}\]

Concept: The Concept of Derivative - Algebra of Derivative of Functions
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.5 | Q 28 | Page 44

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×