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.
$\endgroup$ 113 Answers
$\begingroup$$$ \frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\sec(x+h)) -\ln(\sec(x))}{h} $$
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 $h = 0$
$$ \frac{d}{dx}\ln\sec(x) = \frac{\sin(\frac{2x}{2})}{\cos(x)} $$
$$ \frac{d}{dx}\ln\sec(x) = \frac{\sin(x)}{\cos(x)} $$
$$ \frac{d}{dx}\ln\sec(x) = \tan(x) $$
$\endgroup$ 1 $\begingroup$$\displaystyle \frac{d}{dx} {\ln(\sec x)}=\lim_{h \to 0} \frac{-ln(\frac{\cos(x)\cos(h) - \sin(x)\sin(h)}{\cos(x)})}{h}$
$\displaystyle= \lim_{h \to 0} \frac{-\ln{(\cos h-\tan x \sin h)}}{h}$
$\displaystyle=\lim_{h \to 0} \frac{-\ln[({1- \tan x \tanh)(\cos h)}]}{h}$
$\displaystyle=\lim_{h \to 0} \frac{-\ln({1- \tan x \tanh)}-\ln{\cos h}}{h}$
$\displaystyle=\lim_{h \to 0} -\frac{\ln({1- \tan x \tanh)}}{h}-\lim_{h \to 0} \frac{\ln{\cos h}}{h}$
$\displaystyle=\lim_{h \to 0} {\frac{\ln({1+(-\tan x \tan h))}}{-\tan x \tan h}}{\frac{\tan x \tan h}{h}}-\lim_{h \to 0} {\frac1{\cos h}}{(-\sin h)}$(Using L'hopital Rule, since ${\ln{\cos h}}\to 0, {h}\to 0$)
$\displaystyle=\lim_{\tan x \tan h \to 0} {\frac{\ln({1+(-\tan x \tan h))}}{-\tan x \tan h}} \lim_{h \to 0} {\frac{\tan x \tan h}{h}}+\lim_{h \to 0} {\tan h} \; (\because h \to 0 \implies \tan h \to 0 \implies \tan x \tan h \to 0)$
$\displaystyle=1. \tan x \lim_{h\to 0} \frac{\tan h}{h} +0 \;(\because \lim_{y\to 0} {\ln{(1+y)} \over y}=1)$
$\displaystyle= \tan x \lim_{h\to 0} \sec^2{h}=\tan x . 1$ (Using L'hopital Rule, since ${\tan h}\to 0, {h}\to 0$)
$=\tan x$
$\endgroup$ $\begingroup$I'm not sure if this is in the spirit of the question, but we can also try proving the chain rule in the special case $(\ln \circ \sec)'(a)=\ln'(\sec a) \cdot \sec'(a)$. Note that$$ \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) \, . $$
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