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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Differentiation

NDA 2015 · Mathematics · Differentiation

If s= t^2+1 , then d^2 s dt^2 is equal to

If s=t2+1\mathrm{s}=\sqrt{\mathrm{t}^2+1}, then d2sdt2\frac{\mathrm{d}^2 \mathrm{ s}}{\mathrm{dt}^2} is equal to

Options

  1. A1s\frac{1}{\mathrm{ s}}
  2. B1s2\frac{1}{\mathrm{ s}^2}
  3. C1s3\frac{1}{\mathrm{ s}^3}
  4. D1s4\frac{1}{\mathrm{ s}^4}

Correct answer

C. 1s3\frac{1}{\mathrm{ s}^3}

Step-by-step solution

s=t2+1⇒dsdt=tt2+1\begin{aligned} & \mathrm{s}=\sqrt{\mathrm{t}^2+1} \\ & \Rightarrow \frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{t}}{\sqrt{\mathrm{t}^2+1}}\end{aligned}d2sdt2=1(t2+1)3⇒d2sdt2=1s3\begin{aligned} & \frac{\mathrm{d}^2 \mathrm{ s}}{\mathrm{dt}^2}=\frac{1}{\sqrt{\left(\mathrm{t}^2+1\right)^3}} \\ \Rightarrow & \frac{\mathrm{ d}^2 \mathrm{ s}}{\mathrm{dt}^2}=\frac{1}{\mathrm{ s}^3}\end{aligned}$
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