#### Provide Solution For R.D. Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 6 Maths Textbook Solution.

Answer: $f'\left ( 0 \right )=m$

Hint:  Differentiate f(x) to find f (x). Put x = 0 in f (x).

Given:$f\left ( x \right )=mx+c$

Solution:

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

$\Rightarrow \frac{d}{dx}\left \{ f\left ( x \right ) \right \}=\frac{d}{dx}\left ( mx+c \right )$

$=\frac{d}{dx}\left ( mx \right )+\frac{d}{dx}\left ( c \right )$

\begin{aligned} &{\left[\because \frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right]} \\ &=m \frac{d}{d x}(\mathrm{x})+\frac{d}{d x}(\mathrm{c}) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \end{aligned}

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

$\therefore$ f (x) = m

For f (0), put x = 0 in f (x), then

f (x) = m

$\Rightarrow$ f  (x) is not affected at x = 0, since in f (x) there is no term of x.