How to differentiate Tan x using First principles

Let f(x) = tan x. Then f ' (x) = lim (h->0) [tan(x+h) - tan x] / h

Remember that tan (a + b) = [tan a + tan b] / [1 - tan a * tan b]

f ' (x) = lim (h->0) [(tan x + tan h) / (1 - tan x * tan h) - tan x] / h

Multiply the -tanx by (1 - tan x * tan h)/(1 - tan x * tan h) to get common denominators

f ' (x) = lim (h->0) [(tan x + tan h) / (1 - tan x * tan h) - tanx(1 - tan x * tan h) / (1 - tan x * tan h)] / h
f ' (x) = lim (h->0) [(tan x + tan h - tan x + (tan x)^2 * tan h) / (1 - tan x * tan h)] / h
f ' (x) = lim (h->0) [tan h + (tan x)^2 * tan h] / [h(1 - tan x * tan h)]
f ' (x) = lim (h->0) [tan h (1 + (tan x)^2] / [h (1 - tan x * tan h)]

Note that 1 + (tan x)^2 = (sec x)^2

f ' (x) = lim (h->0) [tan h / h] * [(sec x)^2 / (1 - tan x * tan h)]
f ' (x) = lim (h->0) [tan h / h] * lim (h->0) [(sec x)^2 / (1 - tan x * tan h)]
f ' (x) = 1 * [(sec x)^2 / (1 - tan x * tan 0)]
f ' (x) = 1 * [(sec x)^2 / (1 - tan x * 0)]
f ' (x) = 1 * [(sec x)^2 / (1 - 0)]
f ' (x) = (sec x)^2

In case you are wondering how I found the lim (h->0) tan h / h = 1 here is a proof:

Note the fact the lim (h->0) sin h / h = 1. I will use that in my proof here.

lim (h->0) tan h / h
lim (h->0) (sin h / cos h) / h
lim (h->0) sin h / (h cos h)
lim (h->0) sin h / h * lim (h->0) 1 / cos h
1 * (1 / cos 0)
1 * (1 / 1)
1

Derivative of tanx
=

lim_(h->0) (tan(h+x)-tan(x))/h
Indeterminate form of type 0/0. Applying L'Hospital's rule we have, lim_(h->0) (-tan(x)+tan(h+x))/h = lim_(h->0) (( d(-tan(x)+tan(h+x)))/( dh))/(( dh)/( dh)):
= lim_(h->0) 1/cos(h+x)^2
The limit of a quotient is the quotient of the limits:
= 1/(lim_(h->0) cos(h+x)^2)
Using the power law, write lim_(h->0) cos(h+x)^2 as (lim_(h->0) cos(h+x))^2:
= 1/(lim_(h->0) cos(h+x))^2
Using the continuity of cos(h) at h = x write lim_(h->0) cos(h+x) as cos(lim_(h->0) (h+x)):