How do you differentiate #f(x) = sin(sqrt(arccosx^2)) # using the chain rule?

1 Answer
Jun 21, 2016

Answer:

#-(xcos(sqrt(arccosx^2)))/(sqrt(1-x^4)*sqrt(arccosx^2))#

Explanation:

To differentiate #f(x)# we have to decompose it into functions then differentiate it using chain rule:

Let:
#u(x) =arccosx^2#
#g(x)=sqrt(x)#
Then,
#f(x)=sin(x)#

The derivative of the composite function using chain rule is stated as follows:

#color(blue)((f(g(u(x))))'=f'(g(u(x)))*g'(u(x))*u'(x))#

Let's find the derivative of each function above:

#u'(x)=-1/sqrt(1-(x^2)^2)*2x#
#color(blue)(u'(x)=-1/(sqrt(1-x^4))*2x#

#g'(x)=1/(2sqrt(x))#
Subtituting #x# by #u(x)# we have:
#color(blue)(g'(u(x))=1/(2sqrt(arccosx^2))#

#f'(x)=cos(x)#
Substituting #x# by #g(u(x))# we have to find #color(red)(g(u(x)))#:

#color(red)(g(u(x))=sqrt(arccosx^2))#
So,
#f'(g(u(x)))=cos(g(u(x))#
#color(blue)(f'(g(u(x)))=cos(sqrt(arccosx^2))#

Substituting the calculated derivatives on the above chain rule we have:

#color(blue)((f(g(u(x))))'=f'(g(u(x)))*g'(u(x))*u'(x)#

#=(-2xcos(sqrt(arccosx^2)))/(2sqrt(1-x^4)*sqrt(arccosx^2))#

#color(blue)(=-(xcos(sqrt(arccosx^2)))/(sqrt(1-x^4)*sqrt(arccosx^2)))#