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

Jan 27, 2015

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

It is important to remember which is the derivative of the $y = \ln \left[f \left(x\right)\right]$, which is the derivative of $y = \sqrt{f \left(x\right)}$, which is the derivative of $y = {\left[f \left(x\right)\right]}^{n}$ and the theorem of the function of function derivation.

• The derivative of $y = \ln \left[f \left(x\right)\right]$ is $y ' = \frac{f ' \left(x\right)}{f} \left(x\right)$.
• The derivative of $y = \sqrt{f \left(x\right)}$ is $y ' = \frac{f ' \left(x\right)}{2 \sqrt{f \left(x\right)}}$.
• The derivative of $y = {\left[f \left(x\right)\right]}^{n}$ is $y ' = n {\left[f \left(x\right)\right]}^{n - 1} f ' \left(x\right)$.
• The theorem of the function of function derivation says that the derivative of $y = \left[f \left(g \left(x\right)\right)\right]$ is $y ' = f ' \left(g \left(x\right)\right) g ' \left(x\right)$.

$y ' = \frac{1}{\sqrt{{x}^{2} - 9}} \frac{1}{2 \sqrt{{x}^{2} - 9}} 2 x = \frac{x}{{x}^{2} - 9}$

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

$\ln \left({x}^{n}\right) = n \log \left(x\right)$

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:

$y ' = \left(\frac{1}{2}\right) \frac{1}{{x}^{2} - 9} 2 x = \frac{x}{{x}^{2} - 9}$