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

May 1, 2018

$\frac{1}{x \ln x}$

#### Explanation:

We have:

$\frac{d}{\mathrm{dx}} \left(\ln \left(\ln {x}^{2}\right)\right)$

According to the chain rule, $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(f\right) \cdot \frac{d}{\mathrm{dx}} g \left(x\right)$

Here, $f \left(u\right) = \ln u$ where $u = \ln {x}^{2}$

Since $\frac{d}{\mathrm{du}} \ln u = \frac{1}{u}$, we now have:

$\frac{1}{u} \cdot \frac{d}{\mathrm{dx}} \ln {x}^{2}$

Here, $f \left(v\right) = \ln v$ where $v = {x}^{2}$, so we have:

$\frac{1}{u} \cdot \frac{1}{v} \cdot 2 x$

Since $u = \ln {x}^{2}$ and $v = {x}^{2}$, we have:

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

Since $\ln {x}^{n} = n \ln x$, we get:

$\frac{2 x}{2 {x}^{2} \ln x}$

$\frac{1}{x \ln x}$

May 1, 2018

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

#### Explanation:

We can use chain rule here. We can write $f \left(x\right) = \ln \left(\ln {x}^{2}\right)$ as

$f \left(x\right) = \ln \left(g \left(x\right)\right)$, $g \left(x\right) = \ln \left(h \left(x\right)\right)$ and $h \left(x\right) = {x}^{2}$

then $\frac{\mathrm{df}}{\mathrm{dg}} = \frac{1}{g \left(x\right)}$, $\frac{\mathrm{dg}}{\mathrm{dh}} = \frac{1}{h \left(x\right)}$ and $\frac{\mathrm{dg}}{\mathrm{dh}} = 2 x$

and using chain rule as $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

= $\frac{1}{g \left(x\right)} \times \frac{1}{h \left(x\right)} \times 2 x$

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

= $\frac{2}{x \ln {x}^{2}}$

Hence $\frac{d}{\mathrm{dx}} \ln \left(\ln {x}^{2}\right) = \frac{2}{x \ln {x}^{2}} = \frac{2}{x 2 \ln x} = \frac{1}{x \ln x}$