Question

If $x=e^{-t},y=t^3$ then $\left(\frac{d^{3}y}{d{x}^3}\right){|}_{t=0}=0$ is equal to

• A. $-6$
• B. $0$
• C. $-3$
• D. $6$

Answer 1

The correct option is A $-6$

$\frac{dx}{dt}=-e^{-t},\frac{dy}{dt}={3t}^2$ so $\frac{dy}{dx}=-{3t}^2{e}^t$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-3[2t{e}^t+t^2{e}^t]\frac{dt}{dx}=3[2t+t^2]e^{2t}$
$\Rightarrow \frac{d^{3}y}{{dx}^2}=3[[2+2t]e^{2t}+2[2t+t^2]e^{2t}]\frac{dt}{dx}=-3[2+2t]e^{2t}{e}^t=-6[1+3t+t^2]e^{3t}$
$\therefore$ when $t=0,\frac{d^{3}y}{{dx}^3}=-6$