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

If $y = x^{3} - 8 x + 7$ and $x = f (t)$ and when $t = 0, x = 3$ and $\frac{d y}{d x} = 2$ , then $\frac{d x}{d t}$ at $t = 0$ is

$y = x^{3} - 8 x + 7; x = f (t), t = 0, x = 3 and \frac{d y}{d x} = 2$

$x = f (t) \Rightarrow x = f (0), x = f (t) \Rightarrow \frac{d x}{d t} = f^{1} (t)$

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

$\frac{d y}{d x} = 3. {(f (t))}^{2} f^{1} (t) - 8 f^{1} (t)$

$at t = 0$

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

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

$f^{|} (t) = \frac{2}{19}$

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