Question

ap math find h' (x) for h(x) = ln x(x-1)^3/root of x minus two

Answer 1

To find h'(x), we can use the quotient rule.

Let's first simplify the expression before taking the derivative.

h(x) = ln[x(x - 1)^3 / sqrt(x - 2)]

Using the properties of logarithms, we can rewrite the expression as:

h(x) = ln(x) + ln[(x - 1)^3 / sqrt(x - 2)]

Now, taking the derivative:

h'(x) = (1/x) + [1 / (x - 1)^3] * (3(x - 1)^2) - [1 / (sqrt(x - 2))] * (1 / 2(sqrt(x - 2)))

Simplifying further:

h'(x) = 1/x + 3(x - 1) / (x - 1)^3 - 1 / (2*(x - 2))

Now, we can simplify the expression if needed.

Answer 2

h(x) = ln(x) + ln[(x - 1)^3 / sqrt(x - 2)]
= ln(x) + 3ln(x-1) - 1/2 ln(x-2)
so h'(x) = 1/x + 3/(x-1) - 1/2 * 1/(x-2)
= (7x^2 + 17x - 4)/(2x(x-1)(x-2))

Answer 3

Correct! The simplified expression for h'(x) is:

h'(x) = (7x^2 + 17x - 4)/(2x(x-1)(x-2))