If #f(x) =-e^(2x-1) # and #g(x) = -3sec^2x^2 #, what is #f'(g(x)) #?

1 Answer
Jun 16, 2017

Answer:

#f'(g(x))=1/(6x)e^(2x-1)cotx^2cos^2x^2#

Explanation:

As #f(x)=-e^(2x-1)# and #g(x)=-3sec^2x^2#

#(df)/(dx)=-e^(2x-1)xx2=-2e^(2x-1)#

and#(dg)/(dx)=-3xx2secx^2xxsecx^2tanx^2xx2x#

= #-12xtanx^2sec^2x^2#

Now #f'(g(x))=(df)/(dg)=((df)/(dx))/((dg)/(dx))#

= #(-2e^(2x-1))/(-12xtanx^2sec^2x^2)#

= #e^(2x-1)/(6xtanx^2sec^2x^2)#

= #1/(6x)e^(2x-1)cotx^2cos^2x^2#