Question

How do I find the derivative of the function `y=ln (sqrt(x^2-9))`?

Answer 1

The derivative of this function is: `y'=x/((x^2-9)`.

It is important to remember which is the derivative of the `y=ln[f(x)]`, which is the derivative of `y=sqrt(f(x))`, which is the derivative of `y=[f(x)]^n` and the theorem of the function of function derivation.

• The derivative of `y=ln[f(x)]` is `y'=(f'(x))/f(x)`.
• The derivative of `y=sqrt(f(x))` is `y'=(f'(x))/(2sqrt(f(x)))`.
• The derivative of `y=[f(x)]^n` is `y'=n[f(x)]^(n-1)f'(x)`.
• The theorem of the function of function derivation says that the derivative of `y=[f(g(x))]` is `y'=f'(g(x))g'(x)`.

`y'=1/(sqrt(x^2-9))1/(2sqrt(x^2-9))2x=x/(x^2-9)`

There is another way to do this derivative, rememberig one property of the logarithmic function:

`ln(x^n)=nlog(x)`

so:

`y=ln(sqrt(x^2−9))=ln(x^2-9)^(1/2)=(1/2)log(x^2-9)`

and the derivative of this function is easier:

`y'=(1/2)1/(x^2-9)2x=x/(x^2-9)`