Question

If $x=\frac{1-t}{1+t}$ and $y=\frac{2t}{1+t}$, then $\frac{d^{2}y}{d{x}^2}$ is equal to

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

Answer 1

The correct option is D $0$

Given : $x=\frac{1-t}{1+t}$ and $y=\frac{2t}{1+t}$
We can write
$x+y=1$
Differentiating w.r.t. $x,$ we get
$\Rightarrow 1+\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-1$

Differentiating w.r.t. $x,$ we get
$\therefore \frac{d^{2}y}{d{x}^2}=0$