Question:
If $y=7 x-x^{3}$ and $x$ increases at the rate of 4 units per second, how fast is the slope of the curve changing when $x=2$ ?
Solution:
Here,
$y=7 x-x^{3}$
$\Rightarrow \frac{d y}{d x}=7 x-x^{3}$
Let $s$ be the slope. Then,
$s=7-3 x^{2}$
$\Rightarrow \frac{d s}{d t}=-6 x \frac{d x}{d t}$
$\Rightarrow \frac{d s}{d t}=-6(4)(2)$ $\left[\because x=2\right.$ and $\frac{d x}{d t}=4$ units / sec $]$
$\Rightarrow \frac{d s}{d t}=-48$
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