How do you differentiate $f (x) = \sqrt{sin (4 x)}$ $f(x) = sqrt(sin(4x)$ using the chain rule?

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Answer 1 · 2016-11-03T19:43:34

$f'(x)=(2cos4x)/sqrt(sin4x)$

$f(x)$ is a composite function of three functions

$color(blue)(u(x)=sqrtx and v(x)=sinx and w(x)=4x)$

$f(x)=u(v(w(x)))=u@(v@w)(x)$

differentiation of $f(x)$ or any composite function is determined by applying the chain rule that follows;

$f'(x)=(u@(v@w)(x))'=(u(v@w(x))))'=u'(v@w(x)))xx(v@w(x))'=u'(v@w(x)))xx(v'(w(x))xxw'(x)$

So,
$color(red)(f'(x)=u'(v@w(x))xx(v'(w(x))xxw'(x))$

$u'(x)=1/(2sqrtx)$
$u'(v@w(x))=1/(2sqrt(v(w(x))))=1/(2sqrt(sin4x))$
Then,
$color(red)(u'(v@w(x))=1/(2sqrt(sin4x)))$

$v'(x)=cosx$
$v'(w(x))=cos(w(x))$
Then,
$color(red)(v'(w(x))=cos4x)$

$color(red)(w'(x)=4$

$color(red)(f'(x)=u'(v@w(x))xx(v'(w(x))xxw'(x))$
$f'(x)=1/(2sqrt(sin4x))xxcos4x xx 4$
$f'(x)=(4cos4x)/(2sqrt(sin4x))$

Therefore,
$f'(x)=(2cos4x)/sqrt(sin4x)$

How do you differentiate f(x) = sqrt(sin(4x) f(x)=√sin(4x)f(x) = sqrt(sin(4x) using the chain rule?