Question

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

Answer 1

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.