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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 69 of 72
Marks: +1, -0
4x+5sin⁡x3x+7cos⁡x\frac{4x+5\sin x}{3x+7\cos x}
Solution:  
Let f (x) = 4x+5sin⁡x3x+7cos⁡x\frac{4x+5\sin x}{3x+7\cos x} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} [f (x)] =
(3x+7cos⁡x)(4x+5sin⁡x)′−(4x+5sin⁡x)(3x+7cos⁡x)′(3x+7cos⁡x)2\frac{(3x+7\cos x)(4x+5\sin x)' - (4x+5\sin x)(3x+7\cos x)'}{(3x+7\cos x)^2}
=
(3x+7cos⁡x)(4+5cos⁡x)−(4x+5sin⁡x)(3−7sin⁡x)(3x+7cos⁡x)2\frac{(3x+7\cos x)(4+5\cos x) - (4x+5\sin x)(3-7\sin x)}{(3x+7\cos x)^2}
=
12x+15xcos⁡x+28cos⁡x+35[cos⁡2x+sin⁡2x]−12x+28xsin⁡x−15sin⁡x(3x+7cos⁡x)2\frac{12x+15x\cos x+28\cos x+35\left[\cos^2 x+\sin^2 x\right]-12x+28x\sin x-15\sin x}{(3x+7\cos x)^2}
=
35+15xcos⁡x+28cos⁡x+28xsin⁡x−15sin⁡x(3x+7cos⁡x)2\frac{35+15x\cos x+28\cos x+28x\sin x-15\sin x}{(3x+7\cos x)^2}
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