find the derivative of

1) y=Ln(9x^2-3).e^(7x)

2)y=Ln((6x^3+x^2-x)/(x-1)(x+1))

please help

To find the derivative of both functions, we can use the product rule and the chain rule.

1) y = ln(9x^2 - 3) * e^(7x)
Let's start by using the product rule, which states that the derivative of the product of two functions is the derivative of the first function multiplied by the second function, plus the derivative of the second function multiplied by the first function.

Let u = ln(9x^2 - 3) and v = e^(7x).
Now, let's find the derivatives of u and v.

Derivative of u:
Using the chain rule, we have:
u' = (1 / (9x^2 - 3)) * (18x)

Derivative of v:
Using the chain rule, we have:
v' = 7e^(7x)

Applying the product rule, we get:
y' = u * v' + v * u'
= (ln(9x^2 - 3) * (7e^(7x))) + (e^(7x) * (1 / (9x^2 - 3)) * (18x))
= 7ln(9x^2 - 3)e^(7x) + (18x / (9x^2 - 3))e^(7x)

2) y = ln((6x^3 + x^2 - x) / ((x-1)(x+1)))
To find the derivative of this function, we will use the quotient rule and the chain rule.

Let u = ln(6x^3 + x^2 - x) and v = (x - 1)(x + 1).
Now, let's find the derivatives of u and v.

Derivative of u:
Using the chain rule, we have:
u' = (1 / (6x^3 + x^2 - x)) * (18x^2 + 2x - 1)

Derivative of v:
Using the product rule, we have:
v' = (1 * (x + 1)) + ((x - 1) * 1)
= 2x

Applying the quotient rule, we get:
y' = (u' * v - u * v') / (v^2)
= ((1 / (6x^3 + x^2 - x)) * (18x^2 + 2x - 1) * (x - 1)(x + 1) - ln(6x^3 + x^2 - x) * 2x) / ((x - 1)(x + 1))^2

Simplifying this expression may be challenging, but these steps should help you find the derivative of both functions.