How do we find the derivative of tanx from first principal?

Aug 4, 2017

$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$

Explanation:

According to first principal, if $y = f \left(x\right)$, then

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

Here we have $y = f \left(x\right) = \tan x$

hence $f \left(x + \delta x\right) = \tan \left(x + \delta x\right)$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = L {t}_{\delta x \to 0} \frac{\tan \left(x + \delta x\right) - \tan x}{\delta x}$

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

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

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

$= L {t}_{\delta x \to 0} \left(\sin \frac{\delta x}{\delta x} \times \frac{1}{\cos x \cos \left(x + \delta x\right)}\right)$

$= 1 \times {\sec}^{2} x = {\sec}^{2} x$