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

1 Answer
Nov 3, 2016

Answer:

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

Explanation:

#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)#