Question

Question #fe6b2

Answer 1

`f(x) = ln (x^2-1)^(1/2)`

First, let's remember the following logarithmic rule:

`ln x ^a = a * ln x`

According to this rule, your function can be simplified as follows:

`f(x) = 1/2 ln(x^2 - 1)`

Now, let's use the chain rule:

`f(u) = 1/2 ln(u(x))`

`u(x) = x^2 - 1`

You need to differentiate those functions:

`f(u) = 1/2 ln(u(x)) color(white)(xx)=> f'(u) = 1/2 * 1/(u(x)) = 1 / (2u(x))`

`u(x) = x^2 - 1 color(white)(xxxxx) => u'(x) = 2x`

Now, the only thing left to do is multiply those two derivatives:

`f'(x) = f'(u) * u'(x) = 1/ (2 (x^2-1)) * 2x = (2x) / (2(x^2-1)) = x / (x^2-1)`

Hope that this helped!