Question #8e13d

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Answer 1 · 2016-10-03T13:59:42

$= 4/(4x-5)$

If you're uncomfortable with this, set it up for the chain rule:

so re-write as $f(x)=ln (4x-5) = ln u(x)$

where $u(x) = 4x-5$

so $(df)/dx = (df)/(du) * (du)/dx$

$= 1/u * 4$

$= 4/(4x-5)$

Or if you're more fluent:

$d/dx ln (4x-5)$

$= 1/(4x-5) d/dx (4x-5)$

$= 1/(4x-5) * 4$

$= 4/(4x-5)$

Or in general terms

$d/dx ln (f(x))$

$= ( f'(x))/(f(x))$

$= 4/(4x-5)$