Differentiation Question 357
Question: If $ x=t+\frac{1}{t},y=t-\frac{1}{t}, $ then $ \frac{d^{2}y}{dx^{2}} $ is equal to
Options:
A) $ -4t{{(t^{2}-1)}^{-2}} $
B) $ -4t^{3}{{(t^{2}-1)}^{-3}} $
C) $ (t^{2}+1){{(t^{2}-1)}^{-1}} $
D) $ -4t^{2}{{(t^{2}-1)}^{-2}} $
Show Answer
Answer:
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}=( 1+\frac{2}{t^{2}-1} ) $ and $ \frac{d^{2}y}{dx^{2}}=\frac{d}{dt}( \frac{dy}{dx} ).\frac{dt}{dx} $
$ =2.\frac{-1}{{{(t^{2}-1)}^{2}}}.2t\times \frac{t^{2}}{t^{2}-1}=-\frac{4t^{3}}{{{(t^{2}-1)}^{3}}} $ .
BETA
