differentiate this:
$\displaystyle
f (x) = xsecx
$
$\displaystyle
f '(x) = (1)(secx) + (x)(tanxsecx)
= secx + xtanxsecx
$
help pleaseee?
You have:
$\displaystyle f'(x) = \sec x + x \sec x \tan x$
There is a common $\displaystyle \sec x$ in the equation so factorise it out.
$\displaystyle f'(x) = \sec{x} \left( 1 + x\tan x \right)$
You still get full marks for writing it in your form but it looks neater to have it factorised.