What is dy/dx if y=(3x+e^x)(2x^3-lnx)?

We'll differentiate with respect to x and we'll use the
product rule to differentiate, such as:

(f*g)' = f'*g +
f*g'

Let f = 3x+e^x => f' = 3 +
e^x

Let g = 2x^3 - ln x => g' = 6x^2 -
(1/x)

Now, we'll substitute f,g,f',g' into the formula
above:

[(3x+e^x)(2x^3 - ln x)]' = (3 + e^x)(2x^3 - ln x) +
(3x+e^x)[6x^2 - (1/x)]

We'll remove the
brackets:

[(3x+e^x)(2x^3 - ln x)]' = 6x^3 - 3ln x +
e^x*2x^3 - e^x*ln x + 18x^3 - 3 + e^x*6x^2 -
e^x/x

[(3x+e^x)(2x^3 - ln x)]' = 24x^3 - ln x(3 + e^x) +
e^x(2x^3 + 6x^2 - 1/x) - 3

Therefore, the
derivative of the function is dy/dx = 24x^3 - ln x(3 + e^x) + e^x(2x^3 + 6x^2 - 1/x) -
3.