How do you use the chain rule to differentiate $\frac{1}{ln (4 x)}$ $1/ln(4x)$ ?

Question

How do you use the chain rule to differentiate $\frac{1}{ln (4 x)}$ $1/ln(4x)$ ?

1 Answer

Impact of this question

2172 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-28T14:26:44

Let $u = 4 x$ $u = 4x$ , $v = ln (u) = ln (4 x)$ $v = ln(u) = ln(4x)$ .

Differentiating the above equation yields

$\frac{d u}{d x} = 4$ $frac{du}{dx} = 4$
$\frac{d v}{d u} = \frac{1}{u}$ $frac{dv}{du} = 1/u$

Then,

$\frac{1}{ln (4 x)} = \frac{1}{v}$ $1/ln(4x) = 1/v$

$\frac{d}{d x} (\frac{1}{ln (4 x)}) = \frac{d}{d x} (\frac{1}{v})$ $frac{d}{dx}(1/ln(4x)) = frac{d}{dx}(1/v)$

$= \frac{d}{d v} (\frac{1}{v}) \frac{d v}{d u} \frac{d u}{d x}$ $= frac{d}{dv}(1/v) frac{dv}{du} frac{du}{dx}$

$= (- \frac{1}{v^{2}}) \cdot (\frac{1}{u}) \cdot (4)$ $= (-1/v^2) * (1/u) * (4)$

$= (- \frac{1}{{(ln (4 x))}^{2}}) \cdot (\frac{1}{4 x}) \cdot (4)$ $= (-1/(ln(4x))^2) * (1/{4x}) * (4)$

$= - \frac{1}{x {(ln (4 x))}^{2}}$ $= -1/(x (ln(4x))^2 )$

Science

Math

Humanities

... and beyond

How do you use the chain rule to differentiate $\frac{1}{ln (4 x)}$ $1/ln(4x)$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use the chain rule to differentiate 1ln(4x)1/ln(4x)?

1 Answer

Related questions

Impact of this question

How do you use the chain rule to differentiate $\frac{1}{ln (4 x)}$ $1/ln(4x)$ ?