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

If \({\rm{s}} = \sqrt {{{\rm{t}}^2} + 1} \), then \(\frac{{{{\rm{d}}^2

1.	If \({\rm{s}} = \sqrt {{{\rm{t}}^2} + 1} \), then \(\frac{{{{\rm{d}}^2}{\rm{s}}}}{{{\rm{d}}{{\rm{t}}^2}}}\) is equal to
A.	\(\frac{1}{{\rm{s}}}\)
B.	\(\frac{1}{{{{\rm{s}}^2}}}\)
C.	\(\frac{1}{{{{\rm{s}}^3}}}\)
D.	\(\frac{1}{{{{\rm{s}}^4}}}\)
Answer» D. \(\frac{1}{{{{\rm{s}}^4}}}\)

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

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