# What is the second derivative of f(x)= ln sqrt(x)?

The second derivative is

d^2(lnsqrtx)/dx^2=d/dx(dlnsqrtx/dx)=d/dx(1/2*1/x)= =-1/2*1/x^2

Note that $\ln \sqrt{x} = \ln {x}^{\frac{1}{2}} = \frac{1}{2} \cdot \ln x$

Feb 28, 2016

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

#### Explanation:

We can simplify $f \left(x\right)$ through the rule that

${\log}_{a} \left({b}^{c}\right) = c \cdot {\log}_{a} \left(b\right)$

Here, we see that

$f \left(x\right) = \ln \sqrt{x} = \ln \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} \ln \left(x\right)$

So, to find the derivative of this function, we must know that the derivative of $\ln \left(x\right)$ is $\text{1/} x$, so

$f ' \left(x\right) = \frac{1}{2} \left(\frac{1}{x}\right)$

To differentiate this and find the second derivative, use the power rule, recalling that $\text{1/} x = {x}^{-} 1$.

$f ' ' \left(x\right) = \frac{1}{2} \frac{d}{\mathrm{dx}} \left({x}^{-} 1\right) = \frac{1}{2} \left(- {x}^{-} 2\right) = - \frac{1}{2 {x}^{2}}$