Test Index

NCERT Class XI Mathematics - Limits and Derivatives - Solutions

© examsnet.com
Question : 73 of 72
Marks: +1, -0
xsin⁡nx\frac{x}{\sin^n x}
Solution:  
Let f (x) = xsin⁡nx\frac{x}{\sin^n x} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} [f (x)] = sin⁡nx−xnsin⁡n−1xcos⁡x(sin⁡nx)2\frac{\sin^n x - x n \sin^{n-1} x \cos x}{(\sin^n x)^2}
= (sin⁡x)n−1(sin⁡x−nxcos⁡x)(sin⁡x)2n\frac{(\sin x)^{n-1} (\sin x - n x \cos x)}{(\sin x)^{2n}}
= sin⁡x−nxcos⁡x(sin⁡x)n+1\frac{\sin x - n x \cos x}{(\sin x)^{n+1}}
= sin⁡x−nxcos⁡x(sin⁡x)n+1\frac{\sin x - n x \cos x}{(\sin x)^{n+1}}
© examsnet.com
Go to Question:
Prev Question