Question

If $s=\sqrt{t^2+1}$, then $\frac{d^{2}s}{d{t}^2}$ is equal to

• A. $\frac{1}{s}$
• B. $\frac{1}{s^2}$
• C. $\frac{1}{s^3}$
• D. $\frac{1}{s^4}$

Answer 1

Answer: (C)

$\frac{1}{s^3}$

Given that:

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

Differentiating on both sides w.r.t $t$, we get

$\frac{ds}{dt}=\frac{1}{2\sqrt{t^2+1}}.2t=\frac{t}{\sqrt{t^2+1}}$

Differentiating again both sides w.r.t $t$, we get

$\frac{d^{2}s}{d{t}^2}=\frac{\sqrt{t^2+1}.\left(1\right)-t.\frac{1}{2\sqrt{t^2+1}}.2t}{(\sqrt{t^2+1}{)}^2}$

$=\frac{t^2+1-t^2}{(\sqrt{t^2+1}{)}^2\sqrt{t^2+1}}=\frac{1}{(\sqrt{t^2+1}{)}^3}=\frac{1}{s^3}$

Hence, C is the correct option.