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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 59 of 72
Marks: +1, -0
cos⁡x1+sin⁡x\frac{\cos x}{1+\sin x}
Solution:  
Let f (x) = cos⁡x1+sin⁡x\frac{\cos x}{1+\sin x} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} [f (x)] = (1+sin⁡x)(cos⁡x)′−cos⁡x(1+sin⁡x)′(1+sin⁡x)2\frac{(1+\sin x)(\cos x)' - \cos x(1+\sin x)'}{(1+\sin x)^2}
Since ddx\frac{d}{dx} [f (x)] = (1+sin⁡x)(−sin⁡x)′−cos⁡x(1+sin⁡x)′(0+cos⁡x)2\frac{(1+\sin x)(-\sin x)' - \cos x(1+\sin x)'}{(0+\cos x)^2}
= −sin⁡x−sin⁡2x−cos⁡2x(1+sin⁡x)2\frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2}
= −sin⁡x−−(sin⁡2+cos⁡2x)(1+sin⁡x)2\frac{-\sin x -- (\sin^2{} + \cos^2 x)}{(1+\sin x)^2}
∴ ddx[cos⁡x1+sin⁡x]\frac{d}{dx} \left[ \frac{\cos x}{1+\sin x} \right] = −11+sin⁡x\frac{-1}{1+\sin x}.
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