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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Differentiation

NDA 2021 · Mathematics · Differentiation

What is the derivative of ( x)+ ( x) with respect to x at x=e ?

What is the derivative of sin⁡(ln⁡x)+cos⁡(ln⁡x)\sin (\ln x)+\cos (\ln x) with respect to xx at x=ex=e ?

Options

  1. Acos⁡1−sin⁡1e\frac{\cos 1-\sin 1}{e}
  2. Bsin⁡1−cos⁡1e\frac{\sin 1-\cos 1}{e}
  3. Ccos⁡1+sin⁡1e\frac{\cos 1+\sin 1}{e}
  4. D00

Correct answer

A. cos⁡1−sin⁡1e\frac{\cos 1-\sin 1}{e}

Step-by-step solution

y=sin⁡(ln⁡x)+cos⁡(ln⁡x)dydx=ddx(sin⁡(ln⁡x))+ddx(cos⁡(ln⁡x))=cos⁡(ln⁡x)⋅1x+{−sin⁡(ln⁡x)}⋅1x\begin{aligned} y & =\sin (\ln x)+\cos (\ln x) \\ \frac{d y}{d x} & =\frac{d}{d x}(\sin (\ln x))+\frac{d}{d x}(\cos (\ln x)) \\ & =\cos (\ln x) \cdot \frac{1}{x}+\{-\sin (\ln x)\} \cdot \frac{1}{x}\end{aligned}=cos⁡(ln⁡x)−sin⁡(ln⁡x)xdydx∣x=e=cos⁡(ln⁡e)−sin⁡(ln⁡e)e=cos⁡(1)−sin⁡(1)e\begin{aligned} &= \frac{\cos (\ln x)-\sin (\ln x)}{x} \\ &\left.\frac{d y}{d x}\right|_{x=e}=\frac{\cos (\ln e)-\sin (\ln e)}{e}=\frac{\cos (1)-\sin (1)}{e}\end{aligned}$
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