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

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### 1 Answer

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.**