How do you differentiate  y=ln((6x-5)^6)  using the chain rule?

Jun 1, 2016

$\frac{d}{\mathrm{dx}} \left(\ln \left({\left(6 x - 5\right)}^{6}\right)\right) = \frac{36}{6 x - 5}$

Explanation:

$\frac{d}{\mathrm{dx}} \left(\ln \left({\left(6 x - 5\right)}^{6}\right)\right)$

Applying chain rule,$\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$

Let,${\left(6 x - 5\right)}^{6} = u$
$= \frac{d}{\mathrm{du}} \left(\ln \left(u\right)\right) \frac{d}{\mathrm{dx}} \left({\left(6 x - 5\right)}^{6}\right)$

We know,
$\frac{d}{\mathrm{du}} \left(\ln \left(u\right)\right) = \frac{1}{u}$
and,
$\frac{d}{\mathrm{dx}} \left({\left(6 x - 5\right)}^{6}\right) = 36 {\left(6 x - 5\right)}^{5}$

So,
$= \setminus \frac{1}{u} 36 \setminus {\left(6 x - 5 \setminus\right)}^{5}$

substituting back,$\setminus : u = \setminus {\left(6 x - 5 \setminus\right)}^{6}$

$= \setminus \frac{1}{{\left(6 x - 5\right)}^{6}} 36 {\left(6 x - 5\right)}^{5}$

Simplifying it,we get,

$\frac{36}{6 x - 5}$

Jun 1, 2016

Use properties of logarithms to write $y = 6 \ln \left(6 x - 5\right)$ then us $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$

Explanation:

$y = \ln \left({\left(6 x - 5\right)}^{6}\right) = 6 \ln \left(6 x - 5\right)$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 \left[\frac{1}{6 x - 5} \frac{d}{\mathrm{dx}} \left(6 x - 5\right)\right]$

$= 6 \left[\frac{1}{6 x - 5} \left(6\right)\right]$

$= \frac{36}{6 x - 5}$