`y = ln(sqrt(x^2 - 4))` Find the derivative of the function.

`y=ln(sqrt(x^2-4))`

First, use the formula:

`(lnu)'= 1/u*u'`

Applying that formula, the derivative of the function will be:

`y' =1/sqrt(x^2-4) * (sqrt(x^2-4))'`

To take the derivative of the inner function, express the radical in exponent form.

`y'=1/sqrt(x^2-4)*((x^2-4)^(1/2))'`

Then, use the formula:

`(u^n)'=n*u^(n-1) * u'`

So, y' will become:

`y'=1/sqrt(x^2-4) * 1/2(x^2-4)^(-1/2)*(x^2-4)'`

To take the derivative of the innermost function, use the formulas:

`(x^n)'=n*x^(n-1)`

`(c)' = 0`

Applying these two formulas, y' will become:

`y'=1/sqrt(x^2-4) *1/2(x^2-4)^(-1/2)*(2x-0)`

Simplifying...

`y=ln(sqrt(x^2-4))`

First, use the formula:

`(lnu)'= 1/u*u'`

Applying that formula, the derivative of the function will be:

`y' =1/sqrt(x^2-4) * (sqrt(x^2-4))'`

To take the derivative of the inner function, express the radical in exponent form.

`y'=1/sqrt(x^2-4)*((x^2-4)^(1/2))'`

Then, use the formula:

`(u^n)'=n*u^(n-1) * u'`

So, y' will become:

`y'=1/sqrt(x^2-4) * 1/2(x^2-4)^(-1/2)*(x^2-4)'`

To take the derivative of the innermost function, use the formulas:

`(x^n)'=n*x^(n-1)`

`(c)' = 0`

Applying these two formulas, y' will become:

`y'=1/sqrt(x^2-4) *1/2(x^2-4)^(-1/2)*(2x-0)`

Simplifying it will result to:

`y'=1/sqrt(x^2-4)*1/2(x^2-4)^(-1/2)*2x`

`y'=1/sqrt(x^2-4)*1/2*1/(x^2-4)^(1/2)*2x`

`y'=1/sqrt(x^2-4)*1/2*1/sqrt(x^2-4)*2x`

`y'=x/(x^2-4)`

Therefore, the derivative of the given function is `y'=x/(x^2-4)` .