A) \[-4t{{({{t}^{2}}-1)}^{-2}}\]
B) \[-4{{t}^{3}}{{({{t}^{2}}-1)}^{-3}}\]
C) \[({{t}^{2}}+1){{({{t}^{2}}-1)}^{-1}}\]
D) \[-4{{t}^{2}}{{({{t}^{2}}-1)}^{-2}}\]
Correct Answer: B
Solution :
We have \[\frac{dx}{dt}=1-\frac{1}{{{t}^{2}}},\ \ \frac{dy}{dt}=1+\frac{1}{{{t}^{2}}}\] \[\therefore \frac{dy}{dx}=\frac{{{t}^{2}}+1}{{{t}^{2}}-1}=\left( 1+\frac{2}{{{t}^{2}}-1} \right)\]and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dt}\left( \frac{dy}{dx} \right).\frac{dt}{dx}\] \[=2.\frac{-1}{{{({{t}^{2}}-1)}^{2}}}.2t\times \frac{{{t}^{2}}}{{{t}^{2}}-1}=-\frac{4{{t}^{3}}}{{{({{t}^{2}}-1)}^{3}}}\].
You need to login to perform this action.
You will be redirected in
3 sec
