How do you use the limit definition to find the derivative of #f(x)=2/(5x+1)^3#?

1 Answer
May 29, 2017

#d/dx (2/(5x+1)^3) = - 10 /((5x+1)^4) #

Explanation:

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) #