Question

If $x=5t$ and $y=e^t+3t,$ then $\frac{d^{2}y}{d{x}^2}$ at $t=0$ is equal to

• A. $\frac{1}{24}$
• B. $24$
• C. $25$
• D. $\frac{1}{25}$

Answer 1

The correct option is D $\frac{1}{25}$

We have,
$x=5t$
$\Rightarrow \frac{dx}{dt}=5$
and, $y=e^t+3t$
$\Rightarrow \frac{dy}{dt}=e^t+3$
$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\Rightarrow \frac{dy}{dx}=\frac{e^t+3}{5}=\frac{1}{5}(e^t+3)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{1}{5}(e^t+3)\right)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{5}\frac{d}{dt}(e^t+3)\times \frac{dt}{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{e^t}{5}\times \frac{1}{5}=\frac{e^t}{25}$
At $t=0,$
$\frac{d^{2}y}{d{x}^2}=\frac{1}{25}$