# Differentiate secx using the definition of differential?

Jan 26, 2017

$\frac{d}{\mathrm{dx}} \sec x = \sec x \tan x$

#### Explanation:

For a function $f \left(x\right)$, its differential $\frac{\mathrm{df}}{\mathrm{dx}} = L {t}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

Hence for $\sec x$, $\frac{d}{\mathrm{dx}} \sec x = L {t}_{h \to 0} \frac{\sec \left(x + h\right) - \sec x}{h}$

= $L {t}_{h \to 0} \frac{\frac{1}{\cos} \left(x + h\right) - \frac{1}{\cos} x}{h}$

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

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

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

= $L {t}_{h \to 0} \sin \frac{x + \frac{h}{2}}{\cos \left(x + h\right) \cos x} \times L {t}_{h \to 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$xx

= $\sin \frac{x}{\cos x \cos x} \times 1$

= $\frac{1}{\cos} x \times \sin \frac{x}{\cos} x$

= $\sec x \tan x$