Question

Let $y=x^3-8x+7$ and $x=f\left(t\right)$. If $\frac{dy}{dt}=2$ and $x=3$ at $t=0$, then the value of $\frac{dx}{dt}$ at $t=0$ is

• A. $\frac{3}{19}$
• B. $\frac{2}{19}$
• C. $\frac{5}{19}$
• D. $\frac{7}{19}$

Answer 1

Answer: (B)

$\frac{2}{19}$

we have $y=x^3-8x+7$
Differentiate w.r. to ${}^{\prime }{x}^{\prime }$
$\therefore \frac{dy}{dx}=3x^2-8$
It is given that when $t=0$, $x=3$. Therefore, when $t=0$,
$\frac{dy}{dx}=3\times 3^2-8=19$
Also, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ ........(1)
Since when $t=0$, $\frac{dy}{dx}=19$ and $\frac{dy}{dt}=2$, from (1)
$19=\frac{2}{\frac{dx}{dt}}$
$\therefore \frac{dx}{dt}=\frac{2}{19}$