# How to find f'(0) ?

## Mar 18, 2017

$f ' \left(0\right) = - 6$

#### Explanation:

As $f \left(x\right) = \frac{{x}^{2} + 3}{2 x - 1}$

and using quotient rule,

$f ' \left(x\right) = \frac{2 x \times \left(2 x - 1\right) - 2 \times \left({x}^{2} + 3\right)}{2 x - 1} ^ 2$

= $\frac{4 {x}^{2} - 2 x - 2 {x}^{2} - 6}{2 x - 1} ^ 2$

= $\frac{2 {x}^{2} - 2 x - 6}{2 x - 1} ^ 2$

= $\frac{2 \left({x}^{2} - x - 3\right)}{2 x - 1} ^ 2$

and $f ' \left(0\right) = \frac{2 \left({0}^{2} - 0 - 3\right)}{2 \times 0 - 1} ^ 2 = - \frac{6}{1} = - 6$

Mar 18, 2017

$f ' \left(0\right) = - 6$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + 3}{2 x - 1}$

Using the quotient rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(2 x - 1\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 3\right) - \left({x}^{2} + 3\right) \frac{d}{\mathrm{dx}} \left(2 x - 1\right)}{2 x - 1} ^ 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 x \left(2 x - 1\right) - 2 \left({x}^{2} + 3\right)}{2 x - 1} ^ 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{4 {x}^{2} - 2 x - 2 {x}^{2} - 6}{2 x - 1} ^ 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 {x}^{2} - 2 x - 6}{2 x - 1} ^ 2$

and for $x = 0$

${\left[\frac{\mathrm{df}}{\mathrm{dx}}\right]}_{x = 0} = - 6$

Mar 18, 2017

$f ' \left(0\right) = - 6.$

#### Explanation:

Recall that, $f ' \left(a\right) = {\lim}_{x \to a} \left\{\frac{f \left(x\right) - f \left(a\right)}{x - a}\right\} .$

$\therefore f ' \left(0\right) = {\lim}_{x \to 0} \left\{\frac{f \left(x\right) - f \left(0\right)}{x - 0}\right\} = {\lim}_{x \to 0} \left\{\frac{f \left(x\right) - f \left(0\right)}{x}\right\} .$

$f \left(x\right) = \frac{{x}^{2} + 3}{2 x - 1} \Rightarrow f \left(0\right) = \frac{0 + 3}{0 - 1} = - 3.$

$\therefore f ' \left(0\right) = {\lim}_{x \to 0} \left[\frac{1}{x} \left\{\frac{{x}^{2} + 3}{2 x - 1} - \left(- 3\right)\right\}\right] ,$

$= {\lim}_{x \to 0} \frac{1}{x} \left[\frac{{x}^{2} + 3 + 3 \left(2 x - 1\right)}{2 x - 1}\right] ,$

$= {\lim}_{x \to 0} \frac{1}{x} \left\{\frac{{x}^{2} + 6 x}{2 x - 1}\right\} ,$

$= {\lim}_{x \to 0} \frac{x \left(x + 6\right)}{x \left(2 x - 1\right)} ,$

$= {\lim}_{x \to 0} \left\{\frac{x + 6}{2 x - 1}\right\} .$

$\Rightarrow f ' \left(0\right) = \frac{0 + 6}{0 - 1} = - 6.$

Enjoy Maths.!