By definition the derivative of the function #sqrt(cosx)# is:
#d/dx sqrt(cosx) = lim_(h->0) (sqrt(cos(x+h))-sqrt(cosx))/h#
Multiply numerator and denominator by the quantity::
#(sqrt(cos(x+h))+sqrt(cosx)) !=0# for #h < pi#
#d/dx sqrt(cosx) = lim_(h->0) (sqrt(cos(x+h))-sqrt(cosx))/h*(sqrt(cos(x+h))+sqrt(cosx)) /(sqrt(cos(x+h))+sqrt(cosx)) #
and use the identity: #(a+b)(a-b) = a^2-b^2#
#d/dx sqrt(cosx) = lim_(h->0) (cos(x+h)-cosx)/(h(sqrt(cos(x+h))+sqrt(cosx))#
Use now the trigonometric identity: #cos (alpha+beta) = cos alpha cos beta - sin alpha sin beta#
#d/dx sqrt(cosx)= lim_(h->0) (cosx cos h -sinx sin h -cosx)/(h(sqrt(cos(x+h))+sqrt(cosx))#
#d/dx sqrt(cosx) = lim_(h->0) (-cosx (1-cos h) -sinx sin h )/(h(sqrt(cos(x+h))+sqrt(cosx))#
Separating in two terms:
#d/dx sqrt(cosx) = lim_(h->0) -cosx/(sqrt(cos(x+h))+sqrt(cosx)) (1-cos h)/h +lim_(h->0) -sinx/(sqrt(cos(x+h))+sqrt(cosx)) sin h /h#
We can now use two well known trigonometric limits:
#lim_(h->0) sin h /h = 1#
#lim_(h->0) (1-cos h)/h =0#
so the first term is null, and the second becomes:
#d/dx sqrt(cosx) = - sinx/(2sqrt(cosx))#
