How do you differentiate #x^2(3x+1)^(1/2)#?

1 Answer
May 19, 2015

The answer is : #f'(x) = (x(15x + 4))/(2sqrt(3x+1))#.

Let's have a quick look at your function :

#f(x) = x^2*(3x+1)^(1/2) = g(x)*h(x)#

The derivative of such a form is given by the product rule :

#f'(x) = g'(x)h(x) + g(x)h'(x)#

The derivative of #g(x) = x^2# is #g'(x) = 2x#

and the derivative of #h(x) = (3x+1)^(1/2)# is

#h'(x) = 1/2(3x+1)^(-1/2)*3 = 3/(2(3x+1)^(1/2))#

Therefore, the derivative of #f(x)# is :

#f'(x) = 2x*(3x+1)^(1/2) + x^2*3/(2(3x+1)^(1/2)) = (4x*((3x+1)^(1/2))^2 + 3x^2)/(2(3x+1)^(1/2)) = (4x(3x+1) + 3x^2)/(2(3x+1)^(1/2)) = (15x^2 + 4x)/(2(3x+1)^(1/2)) = (x(15x + 4))/(2sqrt(3x+1))#.

That's it.