Question

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

• 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

Answer: (D)

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