Using the lit definition we have:
#d/dx(2sqrtx-1/(2sqrtx)) = lim_(h->0) (2sqrt(x+h) -1/(2sqrt(x+h)) - 2sqrtx+1/(2sqrtx))/h#
#d/dx(2sqrtx-1/(2sqrtx)) = lim_(h->0) (2sqrt(x+h) - 2sqrtx) /h - (1/(2sqrt(x+h))-1/(2sqrtx))/h#
Now rationalize the numerator of the first term:
#(2sqrt(x+h) - 2sqrtx) /h = (2sqrt(x+h) - 2sqrtx) /h xx (sqrt(x+h) + sqrtx)/ (sqrt(x+h) +sqrtx)#
#(2sqrt(x+h) - 2sqrtx) /h = (2(x+h) - 2x) /(h (sqrt(x+h) +sqrtx)#
#(2sqrt(x+h) - 2sqrtx) /h = (2cancelh) /(cancelh (sqrt(x+h) +sqrtx)#
So:
#lim_(h->0) (2sqrt(x+h) - 2sqrtx) /h = 1/sqrtx#
For the second term of the sum:
#(1/(2sqrt(x+h))-1/(2sqrtx))/h = (2sqrtx -2sqrt(x+h))/(hsqrtxsqrt(x+h))#
and in the same way as above:
#(1/(2sqrt(x+h))-1/(2sqrtx))/h = (2x -2(x+h))/(hsqrtxsqrt(x+h)(sqrtx +sqrt(x+h))#
#(1/(2sqrt(x+h))-1/(2sqrtx))/h = (-2cancelh)/(cancelhsqrtxsqrt(x+h)(sqrtx +sqrt(x+h))#
so that:
#lim_(h->0)(1/(2sqrt(x+h))-1/(2sqrtx))/h = -1/(xsqrtx)#
Finally:
#d/dx(2sqrtx-1/(2sqrtx)) = 1/sqrtx+1/(xsqrtx)#
