Question

If $x= \frac{1-t}{1+t} ; y= \frac{2t}{1+t}, $ then $\frac{d^{2}y}{dx^{2}} = $

• A. $ \frac{2t}{\left(1+t\right)^{2}} $
• B. $ \frac{1}{\left(1+t\right)^{4}} $
• C. $ \frac{2t^{2}}{\left(1+t\right)^{2}} $
• D. $0$

Answer 1

The Correct Option is D

$ x= \frac{1-t}{1+t}, y = \frac{2t}{1+t}$
$ \frac{dx}{dt} = \frac{\left(1+t\right)\left(-1\right) -\left(1-t\right).1}{\left(1+t\right)^{2}} = \frac{-2}{\left(1+t\right)^{2}}$
$ \frac{dy}{dt} = \frac{\left(1+t\right)2-2t.1}{\left(1 +t\right)^{2}} = \frac{2+2t-2t}{\left(1+t\right)^{2}} = \frac{2}{\left(1+t\right)^{2}} $
$\frac{dy}{dx}= \frac{dy}{dt}. \frac{dt}{dx} = \left(\frac{2}{\left(1+t\right)^{2}}\right)\left(\frac{\left(1+t\right)^{2}}{-2}\right) = - 1 $
$\therefore \:\:\: \frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left(-1\right) = 0$