# How do you differentiate y= ln (ln6x)?

##### 1 Answer
Apr 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x \ln \left(6 x\right)}$

#### Explanation:

You will have to use the chain rule twice, as there is one embedded function inside another: $6 x$ inside $\ln$ inside another $\ln$.

The chain rule states that

$f \left(x\right) = g \left(h \left(x\right)\right)$
$f ' \left(x\right) = h ' \left(x\right) g ' \left(h\right)$

Let's solve for the internal $\ln \left(6 x\right)$ first.

$h ' \left(x\right) = \frac{d}{\mathrm{dx}} 6 x = 6$
$g ' \left(h\right) = \frac{1}{h \left(x\right)} = \frac{1}{6 x}$
$\frac{d}{\mathrm{dx}} \ln \left(6 x\right) = \frac{6}{6 x} = \frac{1}{x}$

That is just the inside function. The derivative of the whole thing will need to use the chain rule again.

$f ' \left(x\right) = h ' \left(x\right) g ' \left(h\right)$
$h ' \left(x\right) = \frac{1}{x}$
$g ' \left(h \left(x\right)\right) = \frac{1}{h \left(x\right)} = \frac{1}{\ln} \left(6 x\right)$

$h ' \left(x\right) g ' \left(h\right) = \frac{1}{x} \cdot \frac{1}{\ln} \left(6 x\right) = \frac{1}{x \ln \left(6 x\right)}$