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.

$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\sec(x+h)) -\ln(\sec(x))}{h}
$$
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 $h = 0$
$$
\frac{d}{dx}\ln\sec(x) = \frac{\sin(\frac{2x}{2})}{\cos(x)}
$$
$$
\frac{d}{dx}\ln\sec(x) = \frac{\sin(x)}{\cos(x)}
$$
$$
\frac{d}{dx}\ln\sec(x) = \tan(x)
$$
