Question

If $x=\frac{1+t}{t^3},y=\frac{3}{2t^2}+\frac{2}{t}$, then $x{\left(\frac{dy}{dx}\right)}^3-\frac{dy}{dx}$ is equal to

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

given,

$x=\frac{1+t}{t^3},y=\frac{3}{2t^2}+\frac{2}{t}$

$x=\frac{1}{t^3}+\frac{1}{t^2},y=\frac{3}{2t^2}+\frac{2}{t}$

$\frac{dx}{dt}=\frac{-3}{t^4}-\frac{2}{t^3},\frac{dy}{dt}=\frac{-3}{t^3}-\frac{2}{t^2}$

$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=t$

$x{\left(\frac{dy}{dx}\right)}^3-\frac{dy}{dx}=x{t}^3-t$

$=1(\because x=\frac{1+t}{t^3})$