If #f(x) =sec^2(x/2) # and #g(x) = sqrt(5x-1 #, what is #f'(g(x)) #?

1 Answer
Oct 22, 2017

Answer:

#f'(g (x))=tan ((sqrt (5x-1))/2)xxsec^2 ((sqrt (5x-1))/2)#

Explanation:

#f'(x)=??#
#" "#
#f (x) " "# is a composite function composed of two functions
#" "#
Trigonometric function #u (x)=sec(x/2)# and
#" "#
Polynomial function #v (x)=x^2#
#" "#
#f'(x)# is determined by applying the chain rule .
#" "#
#f (x)=v (u (x))#
#" "#
#f'(x)=v'(u (x)) xx u'(x)#
#" "#
#v'(x)=(x^2)'=2x#
#" "#
#u'(x)=(sec(x/2))'=(x/2)'xxsec'(x/2)#
#" "#
#u'(x)=1/2xxtan(x/2)sec (x/2)#
#" "#
So,#f'(x)# is determined by substituting the differentiated functions
#" "#
#f'(x)=v'(u (x)) xx u'(x)=2 (u (x))xx1/2xxtan(x/2)sec (x/2)#
#" "#
#f'(x)=2sec (x/2)xx1/2xxtan(x/2)sec (x/2)#
#" "#
#f'(x)=tan (x/2)xxsec^2 (x/2)#
#" "#
#f'(g (x))=tan ((g (x))/2)xxsec^2 ((g (x))/2)#
#" "#
#f'(g (x))=tan ((sqrt (5x-1))/2)xxsec^2 ((sqrt (5x-1))/2)#