Given the function:
#y(x) = sin^2x#
Evaluate the incremental ratio:
#(y(x+h)-y(x))/h = (sin^2 (x+h)- sin^2x)/h#
Use now the trigonometric identity:
#sin^2 alpha = (1-cos2alpha)/2#
to get:
#(y(x+h)-y(x))/h = ((1 - cos(2 (x+h))) - (1 - cos 2x))/(2h)#
#(y(x+h)-y(x))/h = -(cos(2 x+2h) - cos 2x)/(2h)#
Using the formula for the cosine of the sum of two angles:
#cos(2x+2h) = cos2x cos 2h -sin 2x sin 2h#
we have:
#(y(x+h)-y(x))/h = -(cos2x cos 2h -sin 2x sin 2h - cos 2x)/(2h)#
#(y(x+h)-y(x))/h = -(cos2x (1-cos 2h) -sin 2x sin 2h )/(2h)#
We can evaluate now the limit for #h->0# keeping in mind the two well known trigonometric limits:
#lim_(alpha->0) (1-cosalpha)/alpha = 0#
#lim_(alpha->0) sin alpha /alpha = 1#
So:
# lim_(h->0) (y(x+h)-y(x))/h = -cos2x lim_(h->0) (1-cos 2h)/(2h) + sin 2x lim_(h->0) (sin 2h)/(2h) #
# lim_(h->0) (y(x+h)-y(x))/h = sin2x#
