Given a curve $y = 7x - x^3$ and $x$ increase at the rate of 2 units per second. The rate at which the slope of the curve is changing, when $x = 5$ is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $-60 \text{ units/s}$ ##
Given,
$ y = 7x - x^3 $
Differentiating both sides w.r.t. $x$, we get
$ \frac{dy}{dx} = 7 - 3x^2 $
Thus, slope, $ s = 7 - 3x^2 $
Now, differentiating w.r.t. '$t$', we get
$ \frac{ds}{dt} = -6x \frac{dx}{dt} $
$ ∴\left. \frac{ds}{dt} \right|_{x = 5} = -6 \times 5 \times 2 $ $ [∵\frac{dx}{dt} = 2 \text{ units/s (given)}] $
$ = -60 \text{ units/s} $