Let \[y={{t}^{10}}+1\]and \[x={{t}^{8}}+1,\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is [UPSEAT 2004]

Let \[y={{t}^{10}}+1\]and \[x={{t}^{8}}+1,\]then \[\frac{{{d}^{2}}y}{d

1.	Let \[y={{t}^{10}}+1\]and \[x={{t}^{8}}+1,\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is [UPSEAT 2004]
A.	\[\frac{5}{2}t\]
B.	\[20{{t}^{8}}\]
C.	\[\frac{5}{16{{t}^{6}}}\]
D.	None of these
Answer» D. None of these

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Let \[y={{t}^{10}}+1\]and \[x={{t}^{8}}+1,\]then \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is [UPSEAT 2004]

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