How do you find the second derivative of ln(sqrtx)?

May 13, 2015

In order to derivate a $\ln$ we must remember its derivation rule: $\mathrm{dl} n f \left(x\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$

Let's just rewrite $\sqrt{x}$ as ${x}^{\frac{1}{2}}$. Like this: $\ln \left({x}^{\frac{1}{2}}\right)$

Now, let's derivate it.

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

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

Now we've found $f ' \left(x\right)$, let's solve the first derivative:

$\frac{\mathrm{dl} n \left({x}^{\frac{1}{2}}\right)}{\mathrm{dx}}$ = $\frac{\frac{1}{2 {x}^{\frac{1}{2}}}}{{x}^{\frac{1}{2}}}$

Simplifying it for exponential rules: $\mathrm{dl} n \left({x}^{\frac{1}{2}}\right) = \frac{1}{2 x}$

Deriving $\left(\frac{1}{2 x}\right)$, again, to obtain your original function's second derivative:

(d²ln(x^(1/2)))/(dx²) = (-1/(2x^(2))) (or, alternatively, $- \frac{1}{2} {x}^{- 2}$)

May 13, 2015

$\sqrt{x} = {x}^{\frac{1}{2}}$, so

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

So the derivative is $\frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2} {x}^{- 1}$

The second derivative, then, is

$- \frac{1}{2} {x}^{-} 2 = - \frac{1}{2 {x}^{2}}$