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

1 Answer
Shura
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.

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