By definition:
#f'(x) = lim_(h->0) (f(x+h)-f(x))/h#
so:
#d/dx (2/(5x+1)^3) = lim_(h->0) 2/h(1/(5(x+h)+1)^3-1/(5x+1)^3)#
regroup the binomials at the denominator
#d/dx (2/(5x+1)^3) = lim_(h->0) 2/h(1/((5x+1)+5h)^3-1/(5x+1)^3)#
Perform the difference:
#d/dx (2/(5x+1)^3) = lim_(h->0) 2/h( ( (5x+1)^3 - ((5x+1)+5h)^3 )/(((5x+1)+5h)^3(5x+1)^3))#
Expand now the power of the second binomial at the numerator:
#d/dx (2/(5x+1)^3) = lim_(h->0) 2/h( ( (5x+1)^3 - (5x+1)^3 - 5h(5x+1)^2 - 25h^2(5x+1) -125h^3 )/(((5x+1)+5h)^3(5x+1)^3))#
The first two terms cancel each other:
#d/dx (2/(5x+1)^3) = lim_(h->0) -2/h( ( 5h(5x+1)^2 + 25h^2(5x+1) +125h^3 )/(((5x+1)+5h)^3(5x+1)^3))#
Simplifying #h# and separating the terms:
#d/dx (2/(5x+1)^3) = lim_(h->0) -( 10(5x+1)^2 )/(((5x+1)+5h)^3(5x+1)^3)-lim_(h->0) -( 50h(5x+1) )/(((5x+1)+5h)^3(5x+1)^3)-lim_(h->0) - ( 250h^2)/(((5x+1)+5h)^3(5x+1)^3)#
The last two limits are zero, so:
#d/dx (2/(5x+1)^3) = -( 10(5x+1)^2 )/((5x+1)^6) = - 10 /((5x+1)^4) #
