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If $y=6x-x^{3}$ and $x$ increases at the rate of $5$ units per second, the rate of change of slope when $x=3 $ is

\[\begin{array}{1 1}(1)-90units/sec&(2)90 units/sec\\(3)180units/sec&(4)-180units/sec\end{array}\]

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$y= 6x-x^3$
$\large\frac{dy}{dt} $$=6. \large\frac{dx}{dt}$$ -3x^2 \large\frac{dx}{dt}$
$\large\frac{dy}{dx} $$=(6-3x^2) \large\frac{dx}{dt}$
Slope $m= \large\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$= 6-3x^2$
$m= \large\frac{dy}{dx} $$=6-3x^2$
$ \large\frac{dm}{dt} $$=-6x. \large\frac{dx}{dt}$
Given $\large\frac{dx}{dt} $$=5 ,x=3$
$\large\frac{dm}{dt} $$=-6(3) .5m =-90 \;units /sec$
Hence 1 is the correct answer.
answered May 19, 2014 by meena.p
 

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