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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Differentiation

NDA 2015 · Mathematics · Differentiation

The derivative of (x+ x) with respect to (x+ x) is

The derivative of ln⁡(x+sin⁡x)\ln (x+\sin x) with respect to (x+cos⁡x)(x+\cos x) is

Options

  1. A1+cos⁡x(x+sin⁡x)(1−sin⁡x)\frac{1+\cos x}{(x+\sin x)(1-\sin x)}
  2. B1−cos⁡x(x+sin⁡x)(1+sin⁡x)\frac{1-\cos x}{(x+\sin x)(1+\sin x)}
  3. C1−cos⁡x(x−sin⁡x)(1+cos⁡x)\frac{1-\cos x}{(x-\sin x)(1+\cos x)}
  4. D1+cos⁡x(x−sin⁡x)(1−cos⁡x)\frac{1+\cos x}{(x-\sin x)(1-\cos x)}

Correct answer

A. 1+cos⁡x(x+sin⁡x)(1−sin⁡x)\frac{1+\cos x}{(x+\sin x)(1-\sin x)}

Step-by-step solution

ln⁡(x+sin⁡x)=ydydx=1(x+sin⁡x)(1+cos⁡x)=(1+cos⁡x)(x+sin⁡x)x+cos⁡x=z( say )dzdx=(1−sin⁡x)\begin{aligned} & \ln (x+\sin x)=y \\ & \frac{d y}{d x}=\frac{1}{(x+\sin x)}(1+\cos x) \\ & =\frac{(1+\cos x)}{(x+\sin x)} \\ & x+\cos x=z(\text{ say }) \\ & \frac{d z}{d x}=(1-\sin x) \end{aligned} (say) derivative of ln⁡(x+sin⁡x)\ln (x+\sin x) w.r.t (x+cos⁡x)(x+\cos x) is dydz=(1+cos⁡x)(x+sin⁡x)(1−sin⁡x)\frac{d y}{d z}=\frac{(1+\cos x)}{(x+\sin x)(1-\sin x)}
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