Originally Posted by

**x5pyd3rx** Excuse the triple posting but, i found something stating that

the derivative of a absolute value is u/|u| *du/dx if i can assume that then my answer would be simple

$\displaystyle \ln|sec5x+cot4x|$

$\displaystyle \frac{d}{dx} = \frac{\frac{sec5x + cot4x * sec5x tan5x - csc^2x}{|sec5x +cot4x|}}{|sec5x + cot4x|}$

This is possible because the derivative of ln states that du/dx = u'/u

so then i would simply just cancel the denomerators of both fractions. and my answer would be the numerator's Numerator.

$\displaystyle frac{sec5x + cot4x * sec5x tan5x - csc^2x}{|sec5x +cot4x|}$