How do you differentiate #g(x) =(2x+1)^(5/2) (4x-1)^(3/4) # using the product rule?

1 Answer
Mar 29, 2017

Answer:

#g'(x) =(2(2x+1)^(3/2) (13x - 1))/(root(4)((4x-1))#

Explanation:

Use a combination of the Power Rule #(u^n)' = n u^(n-1) u'#
and the Product Rule #(uv)' = u v' + v u'#:

Let #u = (2x+1)^(5/2), u' = 5/2(2x+1)^(3/2)(2) = 5(2x+1)^(3/2)#
Let #v = (4x - 1)^(3/4), v' = 3/4(4x - 1)^(-1/4)(4) = 3(4x - 1)^(-1/4)#

#g'(x) = (2x+1)^(5/2) * 3(4x - 1)^(-1/4) + (4x - 1)^(3/4)*5(2x+1)^(3/2)#

Factor remembering #x^(5/2) = x^(3/2)x^(2/2) = x^(3/2)x#

#g'(x) = (2x+1)^(3/2)[(3(2x+1))/(4x-1)^(1/4) + 5 (4x - 1)^(3/4)]#

Find a common denominator for the sum (#x^(3/4)x^(1/4) = x#):

#g'(x) = (2x+1)^(3/2)[(6x+3 +5(4x - 1))/(4x-1)^(1/4)]#

Simplify:
#g'(x) = (2x+1)^(3/2)[(6x+3 +20x - 5)/(4x-1)^(1/4)]#

#g'(x) =((2x+1)^(3/2) (26x - 2))/(4x-1)^(1/4)#

#g'(x) =(2(2x+1)^(3/2) (13x - 1))/(root(4)((4x-1))#