Find the derivative of #tan2x# from first principles?

By definition of the derivative # f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #
So with # f(x) = tan2x # we have;

# f'(x)=lim_(h rarr 0) ( tan(2x+2h) - tan 2x ) / h #

Using # tan A =sinA/cosA # we get

# f'(x) = lim_(h rarr 0) ( sin(2x+2h)/cos(2x+2h) - (sin2x)/(cos2x) ) / h #
# " " = lim_(h rarr 0) 1/h{ sin(2x+2h)/cos(2x+2h) - (sin2x)/(cos2x) } #
# " " = lim_(h rarr 0) 1/h{ (sin(2x+2h)cos2x - cos(2x+2h)sin2x)/(cos(2x+2h)cos2x) } #

Using the identity # sin(A-B) -= sinAcosB - cosBsinA # we get;

# f'(x) = lim_(h rarr 0) 1/h{ (sin(2x+2h-2x))/(cos(2x+2h)cos2x) } #
# " " = lim_(h rarr 0) 1/h{ (sin(2h))/(cos(2x+2h)cos2x) } #
# " " = lim_(h rarr 0) 1/(cos(2x+2h)cos2x) * (sin2h)/h #
# " " = lim_(h rarr 0) 2/(cos(2x+2h)cos2x) * (sin2h)/(2h) #
# " " = lim_(h rarr 0) 2/(cos(2x+2h)cos2x) * lim_(h rarr 0) sin(2h)/(2h) #

We now have to rely on some standard limits:

# lim_(h rarr 0)sin h/h =1 => lim_(h rarr 0) (sin 2h)/(2h) =1 #

And so we get:

# f'(x) = 2/(cos2xcos2x) * 1 #
# " " = 2/(cos^2 2x) #
# " " = 2sec^2 2x #