# How to find the derivative of f(x) = √x − x from first principles.?

Jul 18, 2018

Kindly refer to Explanation.

#### Explanation:

Let, $f \left(x\right) = \sqrt{x} - x$.

Recall that, $f ' \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x} \ldots \ldots \ldots \ldots \ldots \left(\ast\right)$.

With $f \left(x\right) = \sqrt{x} - x , \text{ we have, } f \left(t\right) = \sqrt{t} - t$.

:. {(f(t)-f(x)}=(sqrtt-t)-(sqrtx-x),

$\therefore \text{ for } t \ne x , \frac{f \left(t\right) - f \left(x\right)}{t - x} = \frac{\left(\sqrt{t} - \sqrt{x}\right) - \left(t - x\right)}{t - x}$,

$= \frac{\sqrt{t} - \sqrt{x}}{t - x} - \frac{t - x}{t - x}$,

$= \frac{\left(\sqrt{t} - \sqrt{x}\right)}{{\left(\sqrt{t}\right)}^{2} - {\left(\sqrt{x}\right)}^{2}} - 1$,

$= \frac{\left(\sqrt{t} - \sqrt{x}\right)}{\left(\sqrt{t} - \sqrt{x}\right) \left(\sqrt{t} + \sqrt{x}\right)} - 1$,

$\Rightarrow \frac{f \left(t\right) - f \left(x\right)}{t - x} = \frac{1}{\sqrt{t} + \sqrt{x}} - 1 , \left(t \ne x\right)$.

$\text{Hence, from } \left(\ast\right) , f ' \left(x\right) = {\lim}_{t \to x} \left\{\frac{1}{\sqrt{t} + \sqrt{x}} - 1\right\}$,

$= \frac{1}{\sqrt{x} + \sqrt{x}} - 1$.

$\Rightarrow f ' \left(x\right) = \frac{1}{2 \sqrt{x}} - 1$.