A) \[\frac{5}{2}t\]
B) \[20{{t}^{8}}\]
C) \[\frac{5}{16{{t}^{6}}}\]
D) None of these
Correct Answer: C
Solution :
| [c] We have, \[y={{t}^{10}}+1\therefore \frac{dy}{dt}=10{{t}^{9}}\] |
| and \[x={{t}^{8}}+1\Rightarrow \frac{dx}{dt}=8{{t}^{7}}\] |
| \[\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{10{{t}^{9}}}{8{{t}^{7}}}=\frac{5{{t}^{10}}}{4{{t}^{8}}}=\frac{5(y-1)}{4(x-1)}\,\,\,\,\,\,\,\,\,\,...(i)\]Differentiate (i) with respect to x, we get |
| \[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{5}{4}.\frac{(x-1).\frac{dy}{dx}-(y-1)}{{{(x-1)}^{2}}}\] |
| \[=\frac{5}{4}.\frac{1}{(x-1)}\left[ \frac{dy}{dx}-\frac{(y-1)}{(x-1)} \right]\] |
| \[=\frac{5}{4}.\frac{1}{(x-1)}\left[ \frac{5}{4}.\frac{(y-1)}{(x-1)}-\frac{(y-1)}{(x-1)} \right]\] [Putting (i)] |
| \[=\frac{5}{4}.\frac{(y-1)}{{{(x-1)}^{2}}}\left( \frac{5}{4}-1 \right)=\frac{5}{16}.\frac{{{t}^{10}}}{{{({{t}^{8}})}^{2}}}=\frac{5}{16{{t}^{6}}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
