# Question ae9c3

##### 1 Answer
Sep 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\left(1 + 3 x\right) \ln \left(1 + 3 x\right)}$.

#### Explanation:

Let $\left(1 + 3 x\right) = t , \mathmr{and} , \ln \left(1 + 3 x\right) = \ln t = u$, so that,

y=ln(ln(1+3x)=ln(lnt)=lnu#

Thus, $y$ is a function of $u$, $u$ of $t$, and $t$ of $x$.

Accordingly, by The Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}}$=$\frac{\mathrm{dy}}{\mathrm{du}}$$\frac{\mathrm{du}}{\mathrm{dt}}$$\frac{\mathrm{dt}}{\mathrm{dx}}$.................................$\left(\star\right)$

$y = \ln u \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

$u = \ln t \Rightarrow \frac{\mathrm{du}}{\mathrm{dt}} = \frac{1}{t} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$

$t = 1 + 3 x \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = 3. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(3\right)$

Therefore, by $\left(1\right) , \left(2\right) , \left(3\right) , \mathmr{and} , \left(\star\right)$, we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot \frac{1}{t} \cdot 3 = \frac{3}{t u}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{\left(1 + 3 x\right) \ln \left(1 + 3 x\right)}$.

Enjoy Maths.!