# How do find the derivative using first principles formula?

## $4 x$$\sqrt{x}$

Feb 22, 2018

Derivative is $6 \sqrt{x}$

#### Explanation:

In first principle method we calculate derivative for $y = f \left(x\right)$, using definition of derivative as

$\frac{\mathrm{dy}}{\mathrm{dx}}$ or $\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

Here as $f \left(x\right) = 4 x \sqrt{x}$, $f \left(x + h\right) = 4 \left(x + h\right) \sqrt{x + h}$

and $\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{4 \left(x + h\right) \sqrt{x + h} - 4 x \sqrt{x}}{h}$

= $4 {\lim}_{h \to 0} \frac{\left(\left(x + h\right) \sqrt{x + h} - x \sqrt{x}\right) \left(\left(x + h\right) \sqrt{x + h} + x \sqrt{x}\right)}{h \left(\left(x + h\right) \sqrt{x + h} + x \sqrt{x}\right)}$

= $4 {\lim}_{h \to 0} \frac{{\left(x + h\right)}^{3} - {x}^{3}}{h \left(\left(x + h\right) \sqrt{x + h} + x \sqrt{x}\right)}$

= $4 {\lim}_{h \to 0} \frac{{x}^{3} + 3 h {x}^{2} + 3 {h}^{2} x + {h}^{3} - {x}^{3}}{h \left(\left(x + h\right) \sqrt{x + h} + x \sqrt{x}\right)}$

= $4 {\lim}_{h \to 0} \frac{3 {x}^{2} + 3 h x + {h}^{2}}{\left(x + h\right) \sqrt{x + h} + x \sqrt{x}}$

= $4 \times \frac{3 {x}^{2}}{2 x \sqrt{x}}$

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

= $6 \sqrt{x}$