How to differentiate Tan x using First principles

Since you want to differentiate using first principles, differentiate using the definition of the derivative.

If f(x) = tan(x), then:
f'(x) = lim (h-->0) [tan(x + h) - tan(x)]/h.

Since tan(A + B) = [tan(A) + tan(B)]/[1 - tan(A)tan(B)], we see that:
f'(x) = lim (h-->0) [tan(x + h) - tan(x)]/h
= lim (h-->0) {[tan(x) + tan(h)]/[1 - tan(x)tan(h)] - tan(x)}/h
= lim (h-->0) {tan(x) + tan(h) - tan(x)[1 - tan(x)tan(h)]}/{h[1 - tan(x)tan(h)]}
(Note that I multiplied the numerator and denominator by 1 - tan(x)tan(h).)
= lim (h-->0) [tan(x) + tan(h) - tan(x) + tan^2(x)tan(h)]/{h[1 - tan(x)tan(h)]}
= lim (h-->0) [tan(h) + tan^2(x)tan(h)]/{h[1 - tan(x)tan(h)]}
= lim (h-->0) {tan(h)[1 + tan^2(x)]}/{h[1 - tan(x)tan(h)]}, by factoring out tan(h)
= lim (h-->0) [tan(h)sec^2(x)]/{h[1 - tan(x)tan(h)]}, since sec^2(x) = 1 + tan^2(x)
= lim (h-->0) tan(h)/h * lim (h-->0) sec^2(x)/[1 - tan(x)tan(h)]
= (1)[sec^2(x)/(1 - 0)], since lim (h-->0) tan(h)/h = 1
= sec^2(x).

I hope this helps!

f(x) = tan(x)

f'(x) = lim[h→0] [f(x+h) - f(x)] / ((x+h)-x)
...... = lim[h→0] [tan(x+h) - tan(x)] / h
...... = lim[h→0] [sin(x+h)/cos(x+h) - sinx/cosx] / h
...... = lim[h→0] [(cos(x)sin(x+h) - sin(x)cos(x+h)) / (cos(x)cos(x+h))] / h

...... = lim[h→0] [cos(x) (sin(x)cos(h) + cos(x)sin(h)) - sin(x) (cos(x)cos(h) - sin(x)sin(h))] / [h cos(x) cos(x+h)]

...... = lim[h→0] [sin(x)cos(x)cos(h) + cos²(x)sin(h) - sin(x)cos(x)cos(h) + sin²(x)sin(h)] / [h cos(x) cos(x+h)]

...... = lim[h→0] [sin(h) (cos²(x) + sin²(x))] / [h cos(x) cos(x+h)]
...... = lim[h→0] [sin(h)] / [h cos(x) cos(x+h)]
...... = lim[h→0] sin(h)/h * lim[h→0] 1/(cos(x) cos(x+h))
...... = 1 * 1/(cos(x)cos(x+0))
...... = 1/cos²x
...... = sec²x

d/dx tanx = sec^2 x