How do you differentiate $f (x) = 2 x \cdot sin x \cdot cos x$ $f(x)= 2xsinxcosx$ ?

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Answer 1 · 2015-08-06T22:12:09

$f'(x)=2sinxcosx+2xcos^2x-2xsin^2x$

$f=ghk$ => $f'=g'hk+gh'k+ghk'$

With:
$g=2x$ => $g'=2x$
$h=sinx$ => $h'=cosx$
$k=cosx$ => $k'=-sinx$

We then have:
$f'(x)=2sinxcosx+2xcos^2x-2xsin^2x$

Answer 2 · 2015-08-06T22:17:06

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

$f'(x)=(2x)' cdot (sin(x) cdot cos(x))+2x cdot (sin(x) cdot cos(x))'$

$(2x)'=2$

$(sin(x) cdot cos(x))'=sin(x)' cdot cos(x)+sin(x) cdot cos(x)'$
$=cos(x) cdot cos(x)+sin(x) cdot (-sin(x))$
$=cos^2(x)-sin^2(x)$

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

How do you differentiate f(x)= 2x*sinx*cosxf(x)=2x⋅sinx⋅cosxf(x)= 2x*sinx*cosx?