#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 5 Maths Textbook Solution.

Hint: Differentiate f(x) to find f (x). Put x = 4 in f (x).

Given: $f\left ( x \right )=x^{3}+7x^{2}+8x-9$

Solution:

Differentiating f(x) w.r.t x then,

\begin{aligned} \Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\} &=\frac{d}{d x}\left(\mathrm{x}^{3}+7 \mathrm{x}^{2}+8 \mathrm{x}-9\right) \\ &=\frac{d}{d x}\left(\mathrm{x}^{3}\right)+\frac{d}{d x}\left(7 \mathrm{x}^{2}\right)+\frac{d}{d x}(8 \mathrm{x})+\frac{d}{d x}(-9) \end{aligned}

$\left[\because \frac{d}{d x}\left(a x^{2}+b x+c\right)=\frac{d}{d x}\left(a x^{2}\right)+\frac{d}{d x}(b x)+\frac{d}{d x}(c)\right]$

$=\frac{d}{d x}\left(\mathrm{x}^{3}\right)+7 \frac{d}{d x}\left(\mathrm{x}^{2}\right)+8 \frac{d}{d x}(\mathrm{x})+0\; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right]$

$=3\: x^{3-1}+7\left ( 2x^{2-1} \right )+8\left ( 1x^{1-1} \right )$

$\left [ \because \frac{d}{dx}\left ( constant \right ) =0,\left [ \because \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1} \right ]\right ]$

$\therefore f'\left ( x \right )=3x^{2}+14x+8$

Now, for f  (4), we will put x = 4 in f  (x), then:

$f'\left ( 4 \right )=3\left ( 4 \right )^{2}+14\left ( 4 \right )+8$

$=48+56+8$

$=112$