#### Provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 14

Answer: $\frac{dM}{dt}=(48)$

Hint: Here we use the equation of given curve.

Given: Equation of the curve $y=7x-x^{3}$  and $x$  increases at the rate of 4 units per second

Solution: As the differentiate, the above equation with respect to $x$ , we get the slope of the curve

$\frac{d y}{d x}=\frac{d\left(7 x-x^{3}\right)}{d x}$

$\frac{d y}{d x}=\frac{d(7 x)}{d x}-\frac{d\left(x^{3}\right)}{d x}=7-3 x^{2}$     …….(i)

Suppose $M$  be the slope of the given curve then the above equation $M=7-3x^{2}$

Given $x$  increases at the rate of 4 units per second therefore,

$\frac{d x}{d t}=4 \text { units } / \mathrm{sec}$ ……… eq (ii)

Differentiate the equation of the slope that is equation (ii)

\begin{aligned} &\frac{d M}{d t}=\frac{d\left(7-3 x^{2}\right)}{d t} \\\\ &\frac{d M}{d t}=\frac{d(7)}{d t}-\frac{d\left(3 x^{2}\right)}{d t} \end{aligned}

\begin{aligned} &\frac{d M}{d t}=0-(3 \times 2 x) \frac{d x}{d t} \\\\ &\frac{d M}{d t}=-6 x \times \frac{d x}{d t} \end{aligned}

Let’s put value of   $\frac{d x}{d t}=4 \text { units } / \sec$

$\frac{d M}{d t}=-6 x \times 4$

Let’s put value of $x = 2$(given)

$\frac{d M}{d t}=-6 x \times 4=-6 \times 2 \times 4=-48$

The slope cannot be negative,

Thus, the slope of the curve is changing at the rate of 48 units/sec when $x = 2$