If \[x=t+\frac{1}{t},y=t-\frac{1}{t},\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to

\[({{t}^{2}}+1){{({{t}^{2}}-1)}^{-1}}\]

\[-4{{t}^{3}}{{({{t}^{2}}-1)}^{-3}}\]

\[-4{{t}^{2}}{{({{t}^{2}}-1)}^{-2}}\]

If \[x=t+\frac{1}{t},y=t-\frac{1}{t},\]then \[\frac{{{d}^{2}}y}{d{{x}^

1.	If \[x=t+\frac{1}{t},y=t-\frac{1}{t},\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to
A.	\[-4t{{({{t}^{2}}-1)}^{-2}}\]
B.	\[-4{{t}^{3}}{{({{t}^{2}}-1)}^{-3}}\]
C.	\[({{t}^{2}}+1){{({{t}^{2}}-1)}^{-1}}\]
D.	\[-4{{t}^{2}}{{({{t}^{2}}-1)}^{-2}}\]
Answer» C. \[({{t}^{2}}+1){{({{t}^{2}}-1)}^{-1}}\]

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If \[x=t+\frac{1}{t},y=t-\frac{1}{t},\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to

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