If y=x3−8x+7and x=f(t) and  when  t=0,x=3and dydx=2, then dxdt at t=0 is

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

1. A

$\frac{2}{19}$

2. B

$\frac{2}{18}$

3. C

$\frac{2}{17}$

4. D

$\frac{2}{11}$

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### Solution:

$y={x}^{3}-8x+7;x=f\left(t\right),t=0,x=3\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{dy}{dx}=2$

$x=f\left(t\right)⇒x=f\left(0\right),x=f\left(t\right)⇒\frac{dx}{dt}={f}^{1}\left(t\right)$

$y={\left(f\left(t\right)\right)}^{3}-8f\left(t\right)+7$

$\frac{dy}{dx}=3.{\left(f\left(t\right)\right)}^{2}{f}^{1}\left(t\right)-8{f}^{1}\left(t\right)$

$\text{at\hspace{0.17em}\hspace{0.17em}}t=0$

$2=3{\left(f\left(0\right)\right)}^{2}.{f}^{1}\left(t\right)-8.{f}^{1}\left(t\right)$

$2=3{\left(3\right)}^{2}.{f}^{1}\left(t\right)-8.{f}^{1}\left(t\right)$

${f}^{|}\left(t\right)=\frac{2}{19}$  +91

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