How do you find the derivative of $f(x)=8sinxcosx$ ?

Question

6038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-16T09:51:20

$f'(x)=8(cos^2x-sin^2x)=8cos2x$

This will need the product rule:

$f(x)=uv=>f'(x)=vu'+uv'$

$f(x)=8sinxcosx$

$u=8sinx=>u'=8cosx$

$v=cosx=>v'=-sinx$

$:.f'(x)=8cosxxxcosx+8sinx xx(-sinx)$

$f'(x)=8(cos^2x-sin^2x)=8cos2x$

How do you find the derivative of f(x)=8sinxcosxf(x)=8sinxcosx?