Trigonometric Derivative

$f(x) = \frac{1}{\cos x} + \frac{\sin x}{\cos x}$

$f(x) = \sec x + \tan x$

$f'(x) = \sec x \tan x + \sec^{2} x$

$f'(x) = \sec x(\tan x + \sec x)$

$f'(\frac{\pi}{6}) = \sec \frac{\pi}{6}(\tan \frac{\pi}{6} +\sec \frac{\pi}{6})$

$f'(\frac{\pi}{6}) = \frac{2 \sqrt{3}}{3} (\frac{\sqrt{3}}{3} + \frac{2 \sqrt{3}}{3})$

$= \boxed{2}$

by quotient rule dy/dx=1+sinx(-sinx)-(cosx)(cosx)/cos^2

-sinx+sin^2-cos^2x/cos^2x

_sinx-(1)/cos^2x

_(1+sinx)/cos^2x

1+1/2/3/4=2

Given function is ----

$\large f(x)=\frac{1+\sin x}{\cos x}$

Differentiating both sides w.r.t 'x' --

$\large \implies f'(x)=\frac{\cos x (0+\cos x)-(1+\sin x)(-\sin x)}{\cos^2 x}$

$\large \implies f'(x)=\frac{\cos^2 x + \sin x + \sin^2 x}{\cos^2 x}$

$\large \implies f'(\frac{\pi}{6})=\frac{\cos^2 \frac{\pi}{6} + \sin \frac{\pi}{6} + \sin^2 \frac{\pi}{6}}{\cos^2 \frac{\pi}{6}}$

$\large \implies f'(\frac{\pi}{6})=\frac{(\frac{\sqrt{3}}{2})^2+(\frac{1}{2})+(\frac{1}{2})^2}{(\frac{\sqrt{3}}{2})^2}$

$\large \implies f'(\frac{\pi}{6})=\frac{\frac{3}{4}+\frac{1}{2}+\frac{1}{4}}{\frac{3}{4}}$

$\large \implies f'(\frac{\pi}{6})=\frac{\frac{3}{2}}{\frac{3}{4}}$

$\large \implies f'(\frac{\pi}{6})=\frac{1}{\frac{1}{2}} \implies f'(\frac{\pi}{6})=\boxed{2}$

f(x)=secx+tanx then f'(x)=secx.tanx+(secx)^2 substitute x=30degrees weget f'(30)=(2/3)+(4/3)=2

F(x) = sec x + tanx --> f'(x) = sec x(sec x + tan x) --> f(30) = 2/3^1/2 x (3^1/2) = 2. Answer : 2E-0