Maybe this will help.

The function is a product of two functions. f(x) = (6x^2 + 7); g(x) = (3x - 2 + 5/x)

Method 1: Product Rule

d/dx [f(x)g(x)] = f(x)g'(x) + g(x)f'(x). The chant for this for memorization is "1st times the derivative of the 2nd + the 2nd times the derivative of the first."

(6x^2 + 7) <-- first.

(3x - 2 + 5/x) <-- second

(12x) <--- d/dx(first)

As for taking the derivative of 5/x, recall that 5/x = 5x^(-1). Therefore, from the d/dx(Cx^n) = Cnx^(n-1), you can determine that d/dx(5/x) = 5d/dx(1/x) = 5d/dx[x^(-1)] = 5*-1*x^(-1-1) = -5x^(-2) = -5/(x^2)

With this in mind:

[3 - 5/(x^2)] = d/dx(2nd term).

When I was first learning it, I used to say the chant in my head as I was writing them down. "1st times the derivative of the 2nd + the 2nd times the derivative of the 1st."

(6x^2 + 7)[3 - 5/(x^2)] + (3x - 2 + 5/x)(12x) = 18x^2 - 30 + 21 - 35/(x^2) + 36x^2 - 24x + 60 (when you do the foil method of multiplying polynomials.)

Grouping like terms gives you d/dx[f(x)g(x)] = 54x^2 -24x + 51 - 35/(x^2)

Method 2: Multiplying them out first.

(6x^2 + 7)(3x - 2 + 5/x) = 18x^3 -12x^2 + 30x + 21x - 14 + 35/x.

Combining like terms: f(x)g(x) = 18x^3 - 12x^2 + 51x - 14 + 35/x.

Now you can take the derivative using the d/dx(Cx^n) = Cnx^(n-1) approach because it's one long polynomial.

d/dx[f(x)g(x)] = 3*18x^(3-1) - 2*12^(2-1) + 51*1*x^(1-1) - 14 + 35(-1)x^(-1-1)

d/dx[f(x)g(x)] = 54x^2 - 24x + 51 -35/(x^2).