How do you differentiate $f(x)=sec(4x)/tan(4x)$ ?

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Answer 1 · 2017-11-23T09:46:23

$f'(x)=-4csc(4x)cot(4x)$

$sec(4x)=1/cos(4x)$

So, $f(x)=((1/cos(4x)))/tan(4x)=1/(cos(4x)tan(4x))$

$tan(4x)=sin(4x)/cos(4x)$

$cos(4x)*sin(4x)/cos(4x)=(cos(4x)sin(4x))/cos(4x)$

$(cancel(cos(4x))sin(4x))/cancel(cos(4x))=sin(4x)$

$f(x)=1/sin(4x)=u/v$

$u=1$
$v=sin(4x)$

$u'=0$
$v'=4cos(4x)$

$f'(x)=(vu'-uv')/v^2=>(0sin(4x)-1(4(cos(4x))))/sin^2(4x)$

$=-(4cos(4x))/sin^2(4x)$

$=-4/sin(4x)*cos(4x)/sin(4x)$

$=-4/sin(4x)cot(4x)$

$1/sin(4x)=csc(4x)$

$f'(x)=-4csc(4x)cot(4x)$

How do you differentiate f(x)=sec(4x)/tan(4x)f(x)=sec(4x)/tan(4x)?