If {{s}} = {{{{t}}² + 1}, then {{{{{d}}²{{s}}}}{{{{d}}{{{t}}²}} is equal to

Correct Answer is : 1s3{1}{{{{{s}}³}}s31

Correct Answer is : 1s3{1}{{{{{s}}³}}s31

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UPSC NDA II 2015 Mathematics Paper

Question : 67 of 120

Marks: +1, -0

{\rm{s}} = \sqrt {{{\rm{t}}^2} + 1}

\frac{{{{\rm{d}}^2}{\rm{s}}}}{{{\rm{d}}{{\rm{t}}^2}}}

is equal to

Solution:

Concept:

Differentiate

s = \sqrt{t^2 + 1}

twice using the chain rule and quotient rule to find

\frac{d^2 s}{dt^2}

Explanation:

First, find

\frac{ds}{dt}

s = \sqrt{t^2 + 1} = (t^2 + 1)^{1/2}

\frac{ds}{dt} = \frac{1}{2}(t^2 + 1)^{-1/2} \times 2t = \frac{t}{\sqrt{t^2 + 1}} = \frac{t}{s}

Now compute

\frac{d^2 s}{dt^2}

by differentiating

\frac{t}{s}

using the quotient rule:

\frac{d^2 s}{dt^2} = \frac{1 \cdot s - t \cdot \frac{ds}{dt}}{s^2} = \frac{s - t \cdot \frac{t}{s}}{s^2} = \frac{s - \frac{t^2}{s}}{s^2}

Simplify the numerator:

s - \frac{t^2}{s} = \frac{s^2 - t^2}{s}

Since

s^2 = t^2 + 1

, we have

s^2 - t^2 = 1

Then:

\frac{d^2 s}{dt^2} = \frac{1/s}{s^2} = \frac{1}{s^3}

Answer:

\frac{1}{s^3}

— Option C.

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