Advertisement Remove all ads

X Sin X 1 + Cos X - Mathematics

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

\[\frac{x \sin x}{1 + \cos x}\]

Advertisement Remove all ads

Solution

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

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 11 | Page 44

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×