Question #ae9c3

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Question #ae9c3

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Answer 1 · 2016-09-04T06:33:46

$\frac{d y}{d x} = \frac{3}{(1 + 3 x) ln (1 + 3 x)}$ $dy/dx=3/((1+3x)ln(1+3x))$ .

Explanation:

Let $(1 + 3 x) = t, and, ln (1 + 3 x) = ln t = u$ $(1+3x)=t, and, ln(1+3x)=lnt=u$ , so that,

$y = ln (ln (1 + 3 x) = ln (ln t) = ln u$ $y=ln(ln(1+3x)=ln(lnt)=lnu$

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

Accordingly, by The Chain Rule,

$\frac{d y}{d x}$ $dy/dx$ = $\frac{d y}{d u}$ $(dy)/(du)$ $\frac{d u}{d t}$ $(du)/(dt)$ $\frac{d t}{d x}$ $(dt)/(dx)$ ................................. $(⋆)$ $(star)$

$y=lnu rArr dy/(du)=1/u..........................(1)$

$u=lnt rArr (du)/dt=1/t..............................(2)$

$t=1+3x rArr (dt)/dx=3..............................(3)$

Therefore, by $(1),(2),(3), and, (star)$ , we get,

$dy/dx=1/u*1/t*3=3/(tu)$

$:. dy/dx=3/((1+3x)ln(1+3x))$ .

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