#d/dx (sqrt cot x ) = lim_(h->0) (sqrt cot(x+h) - sqrt(cot x))/h#
Rationalize the numerator using the identity #(a^2-b^2) = (a+b)(a-b)#
#d/dx (sqrt cot x ) = lim_(h->0) ((sqrt cot(x+h) - sqrt(cot x))/h)( (sqrt cot(x+h) + sqrt(cot x))/(sqrt cot(x+h) + sqrt(cot x)))#
#d/dx (sqrt cot x ) = lim_(h->0) ( cot(x+h) - (cot x))/(h (sqrt cot(x+h) + sqrt(cot x))#
Now use the trigonometric formulas for the sum of angles:
# cot(x+h) = cos(x+h)/sin(x+h) = (cosxcos h-sinxsin h )/ ( cosx sin h + sinx cos h )#
divide numerator and denominator by #sinx sin h #
# cot(x+h) = (cotxcot h-1 )/ ( cotx +cot h )#
end evaluate the difference:
# cot(x+h) - cotx = (cotxcot h-1 )/ ( cotx +cot h ) - cotx#
# cot(x+h) - cotx = ((cotxcot h-1 ) -cotx( cotx +cot h ))/ ( cotx +cot h ) #
# cot(x+h) - cotx = ((cancel(cotxcot h) -1 -cot^2x- cancel(cotx cot h )))/ ( cotx +cot h ) #
# cot(x+h) - cotx =- ( 1 +cot^2x)/ ( cotx +cot h ) #
and as:
#1+cot^2x = 1+ cos^2x/sin^2x = (sin^2x+cos^2x)/sin^2x = 1/sin^2x = csc^2x#
we have finally:
# cot(x+h) - cotx = - csc^2x/(cotx +cot h )#
So:
#d/dx (sqrt cot x ) = lim_(h->0) ( - csc^2x/(cotx +cot h ))1/(h (sqrt cot(x+h) + sqrt(cot x))#
#d/dx (sqrt cot x ) = lim_(h->0) (- csc^2x/ (sqrt cot(x+h) + sqrt(cot x)))(1/(h(cotx +coth)))#
Now note that:
#lim_(h->0) 1/(h(cotx +coth)) = lim_(h->0) 1/(hcotx + cosh/(sinh/h)) =1#
and we can conclude:
#d/dx (sqrt cot x ) = -csc^2x/(2sqrt(cotx))#
