A) \[\frac{5}{2}t\]
B) \[20{{t}^{8}}\]
C) \[\frac{5}{16{{t}^{6}}}\]
D) None of these
Correct Answer: C
Solution :
Here \[y={{t}^{10}}+1\] and \[x={{t}^{8}}+1\] \[\therefore \] \[{{t}^{8}}=x-1\] \[\Rightarrow \,\,\,{{t}^{2}}={{(x-1)}^{1/4}}\] So, \[y={{(x-1)}^{5/4}}+1\] Differentiate both sides w.r.t. x, \[\frac{dy}{dx}=\frac{5}{4}{{(x-1)}^{1/4}}\] Again, differentiate both sides w.r.t. x, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{5}{16}{{(x-1)}^{-3/4}}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{5}{16{{(x-1)}^{3/4}}}=\frac{5}{16{{({{t}^{2}})}^{3}}}=\frac{5}{16\,{{t}^{6}}}.\]
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