find the derivative of F(x)=x^2-4/1+lnx
To find the derivative of the function F(x) = (x^2 - 4) / (1 + ln(x)), we can use the quotient rule. The quotient rule states that if we have a function in the form f(x)/g(x), then the derivative is given by (g(x)f'(x) - f(x)g'(x)) / [g(x)]^2.
Let's break down the function F(x) = (x^2 - 4) / (1 + ln(x)).
First, let's find the derivative of the numerator, (x^2 - 4). The derivative of x^2 is 2x, and the derivative of -4 is 0. So, the derivative of the numerator, f'(x), is 2x.
Second, let's find the derivative of the denominator, (1 + ln(x)). The derivative of 1 is 0, and the derivative of ln(x) is 1/x. So, the derivative of the denominator, g'(x), is 1/x.
Now, we can substitute these values into the quotient rule formula:
f'(x) = 2x and g'(x) = 1/x
Using the quotient rule, the derivative of F(x) = (x^2 - 4) / (1 + ln(x)) is:
[F'(x)] / [G(x)]^2 = [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2
Replacing f'(x) with 2x, f(x) with (x^2 - 4), g'(x) with 1/x, and g(x) with (1 + ln(x)), we get:
= [(1 + ln(x))(2x) - (x^2 - 4)(1/x)] / [(1 + ln(x))^2]
Simplifying this expression gives us the final answer.