# What is the derivative of x^2/5?

Sep 14, 2016

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2}{5} x$

#### Explanation:

AS $f \left(x\right) = {x}^{2} / 5$

$f \left(x + h\right) = {\left(x + h\right)}^{2} / 5 = \frac{{x}^{2} + 2 h x + {h}^{2}}{5} = {x}^{2} / 5 + \frac{2}{5} h x + {h}^{2} / 5$

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = L {t}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

= $L {t}_{h \to 0} \frac{{x}^{2} / 5 + \frac{2}{5} h x + {h}^{2} / 5 - {x}^{2} / 5}{h}$

= $L {t}_{h \to 0} \frac{\frac{2}{5} h x + {h}^{2} / 5}{h}$

= $L {t}_{h \to 0} \left(\frac{2}{5} x + \frac{h}{5}\right)$

= $\frac{2}{5} x$