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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 60 of 72
Marks: +1, -0
sin⁡x+cos⁡xsin⁡x−cos⁡x\frac{\sin x+\cos x}{\sin x-\cos x}
Solution:  
Let f (x) = sin⁡x+cos⁡xsin⁡x−cos⁡x\frac{\sin x+\cos x}{\sin x-\cos x} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} (f (x)) =
(sin⁡x−cos⁡x)(sin⁡x+cos⁡x)′−(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)′(sin⁡x−cos⁡x)2\frac{(\sin x-\cos x)(\sin x+\cos x)'-(\sin x+\cos x)(\sin x-\cos x)'}{(\sin x-\cos x)^2}
=
(sin⁡x−cos⁡x)(cos⁡x−sin⁡x)−(sin⁡x+cos⁡x)(cos⁡x+sin⁡x)(sin⁡x−cos⁡x)2\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}
=
sin⁡xcos⁡x−sin⁡2x−cos⁡2x+cos⁡xsin⁡x−(sin⁡x+cos⁡x)2(sin⁡x−cos⁡x)2\frac{\sin x \cos x - \sin^2 x - \cos^2 x + \cos x \sin x - (\sin x+\cos x)^2}{(\sin x-\cos x)^2}
= −2(sin⁡x−cos⁡x)2\frac{-2}{(\sin x-\cos x)^2}
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