How do you find the first and second derivative of lnx^(1/2)?

Jun 14, 2016

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

Explanation:

We use the concept of function of a function according to which,.

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$

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

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

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