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Mathematics Question on Continuity and differentiability

If x=1−t1+t;y=2t1+t,x= \frac{1-t}{1+t} ; y= \frac{2t}{1+t}, then d2ydx2=\frac{d^{2}y}{dx^{2}} =

A

2t(1+t)2\frac{2t}{\left(1+t\right)^{2}}

B

1(1+t)4\frac{1}{\left(1+t\right)^{4}}

C

2t2(1+t)2\frac{2t^{2}}{\left(1+t\right)^{2}}

D

00

Answer

00

Explanation

Solution

x=1−t1+t,y=2t1+t x= \frac{1-t}{1+t}, y = \frac{2t}{1+t}
dxdt=(1+t)(−1)−(1−t).1(1+t)2=−2(1+t)2\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}}
dydt=(1+t)2−2t.1(1+t)2=2+2t−2t(1+t)2=2(1+t)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}}
dydx=dydt.dtdx=(2(1+t)2)((1+t)2−2)=−1\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
∴   d2ydx2=ddx(−1)=0\therefore \:\:\: \frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left(-1\right) = 0