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!