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

Question

Luke Alvarez · Accepted Answer

Product rule:
if #f(x) = g(x) * h(x)#, then #f'(x) = g'(x) * h(x) + h'(x) * g(x)#. Here, #g(x) = 5(3x^2 + 1) ^(1/2)# and #h(x) = 3x+1#. Now you need to differentiate #g(x)# and #h(x)#. Let's start with the easy one:
#h(x) = 3x+ 1#
#h'(x) = 3# #g(x)# is the complicated one because there, you will need the chain rule.
#g(x) = 5(3x^2 + 1) ^(1/2) = 5u(x)^(1/2)# where #u(x) = 3x^2 + 1#. With the chain rule, 
#g'(x) = 5*(1/2)* u(x)^(-1/2) * u'(x)#
# = 5/2 * (3x^2 + 1) ^(-1/2) * 6x# 
#= 15x * (3x^2 + 1) ^(-1/2)#
# = (15x) / (3x^2+1)^(1/2)# So, we have
#g(x) = 5(3x^2 + 1) ^(1/2)#
#g'(x) = 15x * (3x^2 + 1) ^(-1/2)# Last thing left to do is to apply the product rule: #f'(x) = g'(x) * h(x) + h'(x) * g(x) #
#= 15x * (3x^2 + 1) ^(-1/2) * (3x+ 1) + 3 * 5(3x^2 + 1) ^(1/2)#
# = (15x(3x+1)) / sqrt(3x^2 + 1) + 15 sqrt(3x^2 + 1)#
# = (15x(3x+1)) / sqrt(3x^2 + 1) + (15 sqrt(3x^2 + 1) * sqrt(3x^2 + 1)) / sqrt(3x^2 + 1)#
#= (45 x^2 + 15x + 15(3x^2 + 1)) / sqrt(3x^2 + 1)#
# = (90 x^2 + 15x + 15) * (3x^2 + 1) ^(1/2)#
# =15 (6 x^2 + x + 1) * (3x^2 + 1) ^(1/2)# I hope that this helped!

Scarlett Allison · Answer

undefined To differentiate $ f(x) = 5(3x^2 + 1)^{\frac{1}{2}} (3x + 1) $ using the product rule, you would differentiate each part separately and then apply the product rule:

1. Differentiate $ 5(3x^2 + 1)^{\frac{1}{2}} $:
   $$ \frac{d}{dx} [5(3x^2 + 1)^{\frac{1}{2}}] = \frac{d}{dx} [5(3x^2 + 1)^{\frac{1}{2}}] = \frac{5}{2} (3x^2 + 1)^{\frac{1}{2} - 1} (6x) = \frac{15x}{\sqrt{3x^2 + 1}} $$

2. Differentiate $ (3x + 1) $:
   $$ \frac{d}{dx} (3x + 1) = 3 $$

3. Apply the product rule:
   $$ f'(x) = \frac{15x}{\sqrt{3x^2 + 1}} (3x + 1) + 5(3x^2 + 1)^{\frac{1}{2}} \cdot 3 $$

Simplify the expression if necessary.