Answered: "V=ln⁡[ln⁡(3x−1)] Find dVdx"

lwfrgin

Answered question

2020-12-12

$V = \ln [\ln (3 x - 1)]$
Find $\frac{d V}{d x}$

Answer & Explanation

FieniChoonin

Skilled2020-12-13Added 102 answers

It is given that
$v = \ln (\ln (3 x - 1))$
We have to find $\frac{d v}{d x}$ .
Let us consider $u = \ln (3 x - 1)$ . Then we have $v = \ln (u)$ . Clearly v is a function of u and u is a function of x. Thus by chain rule we have
$\frac{d v}{d x} = \frac{d v}{d u} \cdot \frac{d u}{d x}$
It is easy to note that
$\frac{d v}{d u} = \frac{d}{d u} (\ln (u) = \frac{1}{u})$
$\frac{d u}{d x} = \frac{d}{d x} (\ln (3 x - 1))$ $\frac{d}{d x} (\ln (y)) = \frac{d}{d y} (\ln (y)) \times \frac{d y}{d x} = (\frac{1}{y}) \times 3 = \frac{3}{3 x - 1}$
Therefore by (Eq-1) we get
$\frac{d v}{d x} = (\frac{1}{u}) \cdot \frac{3}{3 x - 1}$
On substitute back $u = \ln (3 x - 1)$ we get
$\frac{d v}{d x} = \frac{3}{(3 x - 1) \ln (3 x - 1)}$

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