Question

If $x=t^3+t+5$ & $y=\sin t$ then $\frac{d^{2}y}{d{x}^2}=$

• A. $-\frac{(3t^2+1)\sin t+6t\cos t}{(3t^2+1{)}^3}$
• B. $\frac{(3t^2+1)\sin t+6t\cos t}{(3t^2+1{)}^2}$
• C. $-\frac{(3t^2+1)\sin t+6t\cos t}{(3t^2+1{)}^2}$
• D. $\frac{cost}{3t^2+1}$

Answer 1

The correct option is C $-\frac{(3t^2+1)\sin t+6t\cos t}{(3t^2+1{)}^2}$

$x=t^3+t+5$ & $y=\sin t$
Differentiate x and y w.r.t. 't'
$\frac{dx}{dt}=\frac{d}{dt}(t^3+t+5)$
$=3t^2+1$ ..........(1)
$\frac{dy}{dt}=\frac{d}{dt}\sin t$
$=\cos t$ ..........(2)
From 1 and 2
$\frac{dy}{dx}=\frac{\cos t}{3t^2+1}$
$\frac{d^{2}y}{d{x}^2}=\frac{(3t^2+1)\frac{d}{dx}\cos t-\cos t\frac{d}{dx}(3t^2+1)}{{(3t^2+1)}^2}$
$=-\frac{((3t^2+1)\sin t+6t.\cos t)}{{(3t^2+1)}^2}$