Find the derivative of #sin^2x# using first principles?

It is tedious (and space-consuming) to use #lim_(hrarr0)# on every line, so I hope you'll excuse my approach. We'll simplify the difference quotient first, then find the limit.

# ((sin(x+h))^2-(sinx)^2)/h = ((sinxcos h +cos x sin h)^2-(sinx)^2)/h#

# = (color(red)(sin^2xcos^2h)+2sinxcos h cosx sin h +color(blue)(cos^2xsin^2h)-color(red)(sin^2x))/h#

# = (sin^2x(cos^2h-1))/h +(2sinxcos h cosx sin h)/h +(cos^2xsin^2h)/h#

# = sin^2x(cosh-1)/h(cosh+1) +2sinxcos h cosx sin h /h+cos^2xsin h/h sin h#

Taking limit as #h rarr0#, we get

#sin^2(0)(0)(cos 0 +1) +2sinx (cos 0) cosx (1) +cos^0 (1) sin0#

# = 0+2sinx cos x +0#

# = 2sinx cos x#

The definition of the derivative of #y=f(x)# is

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

So with # f(x) = sin^2x # then;

# f'(x)=lim_(h rarr 0) {sin^2(x+h) -sin^2(x)}/h#

Let us focus on the numerator #f(x+h)-f(x)#;

# f(x+h)-f(x) #
# \ \ = sin^2(x+h) -sin^2(x)#
# \ \ = (sinxcos h+cosxsin h)^2 -sin^2(x)#
# \ \ = (sinxcos h)^2+2sinxcos hcosxsin h + (cosxsin h)^2 -sin^2(x)#
# \ \ = sin^2xcos^2h+2sinxcos hcosxsin h + cos^2xsin^2h -sin^2x#
# \ \ = sin^2xcos^2h+2sinxcos hcosxsin h + cos^2xsin^2h -sin^2x#
# \ \ = sin^2x(cos^2h-1)+2sinxcos hcosxsin h + cos^2xsin^2h #
# \ \ = sin^2xsin^2h+2sinxcos hcosxsin h + cos^2xsin^2h #
# \ \ = sin^2h(sin^2x+cos^2x) + 2sinxcos hcosxsin h #
# \ \ = sin^2h+ 2sinxcos hcosxsin h #

And so the limit becomes:

# f'(x)=lim_(h rarr 0) {sin^2h+ 2sinxcos hcosxsin h}/h#
# " "=lim_(h rarr 0) {sin^2h/h+ (2sinxcos hcosxsin h)/h}#
# " "=lim_(h rarr 0) {sin h * sin h/h+ 2sinxcosx*cos h*sin h/h}#

And then using the Fundamental trigonometric calculus limits:

# lim_(theta rarr0) sin theta /theta = 1 #

we have:

# f'(x)= 0 * 1+ 2sinxcosx*1*1#
# " "=2sinxcosx #