# How do you find the first and second derivative of ln(ln x^2)?

Nov 5, 2016

For the first, using chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
For the second, using quotient and product rules.

#### Explanation:

Renaming $v = \ln u$ and $u = {x}^{2}$, we get that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{v} \frac{1}{u} 2 x = \frac{2 x}{u \cdot \ln u} = \frac{2 \cancel{x}}{{x}^{\cancel{2}} \cdot \ln {x}^{2}}$

For the second derivative, we must recall quotient rule...

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

...and product rule (to derive the quotient):

$\left(a \left(x\right) b \left(x\right)\right) ' = a ' b + a b '$

Thus:

$\frac{{\mathrm{dy}}^{2}}{{\mathrm{dx}}^{2}} = \frac{\cancel{0 \cdot \left(x \ln {x}^{2}\right)} - 2 \left(\ln {x}^{2} + 1\right)}{x \ln {x}^{2}} ^ 2$

$\frac{{\mathrm{dy}}^{2}}{{\mathrm{dx}}^{2}} = - \frac{2 \ln {x}^{2} + 2}{{x}^{2} \cdot {\left(\ln {x}^{2}\right)}^{2}}$