Using first principles find derivative of ln(sec(x)) The question is to use first principles only.
Thus I started with the same and got
$$
y = \ln(\sec(x))
$$
$$
\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln(\sec(x+h)) - \ln(\sec(x))}{h}
$$
after this I do not understand how do I eliminate the $h$ in the denominator. I tried to implement $\ln(A) - \ln(B) = \ln\bigl(\frac{A}{B}\bigr)$ which ultimately led to
$$
\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln\bigl(\frac{\sec(x+h)}{\sec(x)}\bigr)}{h}
$$
here I converted $\sec()$ to $\cos()$
$$
\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln\bigl(\frac{\cos(x)}{\cos(x+h)}\bigr)}{h}
$$
Still I cannot proceed further.
 A: $$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\sec(x+h)) -\ln(\sec(x))}{h}
$$
Using $\ln(A) - \ln(B) = \ln(\frac{A}{B})$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\frac{\sec(x+h)}{\sec(x)})}{h}
$$
coverting $\sec(x)$ to $\cos(x)$ using $\cos(x) = \frac{1}{\sec(x)}$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\frac{\cos(x)}{\cos(x+h)})}{h}
$$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x)}{\cos(x+h)}-1)}{h}
$$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{h}
$$
multiplying and dividing by $\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{h}\frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}
$$
repositioning
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}\frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
seperating limit
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}\lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
As $h$ approaches 0 so does $\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$ as the numerator beging to approach $0$. ($\cos(x) - \cos(x)$)
let us assume $t = \frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$ and hence t approaches $0$ when $h$ approaches $0$
hence equation turns out to be
$$
\frac{d}{dx}\ln\sec(x) = \lim_{t\to0} \frac{\ln(1+t)}{t} \lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
Using the standard limit $\lim_{x\to0} \frac{ln(x+1)}{x} = 1$
Therefore
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\cos(x)-\cos(x+h)}{h\cos(x+h)}
$$
Applying $\cos(A) - \cos(B) = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{-2\sin(\frac{2x+h}{2})\sin(\frac{-h}{2})}{h\cos(x+h)}
$$
Bringing the $-2$ down
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})\sin(\frac{-h}{2})}{\frac{-h}{2}\cos(x+h)}
$$
rearranging
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})}{\cos(x+h)} \lim_{h\to0} \frac{\sin(\frac{-h}{2})}{\frac{-h}{2}}
$$
Using standard limit $\lim_{x\to0} \frac{sin(x)}{x} = 1$
Therefore
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})}{\cos(x+h)}
$$
Putting $h = 0$
$$
\frac{d}{dx}\ln\sec(x) = \frac{\sin(\frac{2x}{2})}{\cos(x)}
$$
$$
\frac{d}{dx}\ln\sec(x) = \frac{\sin(x)}{\cos(x)}
$$
$$
\frac{d}{dx}\ln\sec(x) = \tan(x)
$$
A: $\displaystyle \frac{d}{dx} {\ln(\sec x)}=\lim_{h \to 0} \frac{-ln(\frac{\cos(x)\cos(h) - \sin(x)\sin(h)}{\cos(x)})}{h}$
$\displaystyle= \lim_{h \to 0} \frac{-\ln{(\cos h-\tan x \sin h)}}{h}$
$\displaystyle=\lim_{h \to 0} \frac{-\ln[({1- \tan x \tanh)(\cos h)}]}{h}$
$\displaystyle=\lim_{h \to 0} \frac{-\ln({1- \tan x \tanh)}-\ln{\cos h}}{h}$
$\displaystyle=\lim_{h \to 0} -\frac{\ln({1- \tan x \tanh)}}{h}-\lim_{h \to 0} \frac{\ln{\cos h}}{h}$
$\displaystyle=\lim_{h \to 0} {\frac{\ln({1+(-\tan x \tan h))}}{-\tan x \tan h}}{\frac{\tan x \tan h}{h}}-\lim_{h \to 0} {\frac1{\cos h}}{(-\sin h)}$
(Using L'hopital Rule, since ${\ln{\cos h}}\to 0, {h}\to 0$)
$\displaystyle=\lim_{\tan x \tan h \to 0} {\frac{\ln({1+(-\tan x \tan h))}}{-\tan x \tan h}} \lim_{h \to 0} {\frac{\tan x \tan h}{h}}+\lim_{h \to 0} {\tan h} \; (\because h \to 0 \implies \tan h \to 0 \implies \tan x \tan h \to 0)$
$\displaystyle=1. \tan x \lim_{h\to 0}  \frac{\tan h}{h} +0 \;(\because \lim_{y\to 0} {\ln{(1+y)} \over y}=1)$
$\displaystyle= \tan x \lim_{h\to 0} \sec^2{h}=\tan x . 1$ (Using L'hopital Rule, since ${\tan h}\to 0, {h}\to 0$)
$=\tan x$
A: I'm not sure if this is in the spirit of the question, but we can also try proving the chain rule in the special case $(\ln \circ \sec)'(a)=\ln'(\sec a) \cdot \sec'(a)$. Note that
$$
\lim_{x \to a}\frac{\ln(\sec x)-\ln(\sec a)}{x-a}=\lim_{x \to a}\frac{\ln(\sec x)-\ln(\sec a)}{\sec x-\sec a} \cdot \lim_{x \to a}\frac{\sec x-\sec a}{x-a} \label{*}\tag{*} \, .
$$
For the first limit on the RHS of $\eqref{*}$, we can make the substitution $u=\sec x$. As $x\to a$, $u\to\sec a$, and so
$$
\lim_{u \to \sec a}\frac{\ln(u)-\ln(\sec a)}{u-\sec a}=\ln'(\sec a)=\frac{1}{\sec a}
$$
Since the second limit on the RHS is $\sec'(a)=\sec(a)\tan(a)$, we get that
$$
(\ln \circ \sec)'(a)=\tan(a) \, .
$$
