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

1 Answer
Mar 8, 2016

Answer:

#(2.e^(4x) .cose^(4x))/(sqrt(sine^(4x))#

Explanation:

differentiate using the#color(blue)" chain rule "#

#d/dx[f(g(x))] = f'(g(x)) . g'(x) #

This procedure will require to be applied 3 times.

write #f(x) " as " (sine^(4x))^(1/2) #

f'(x)# = 1/2(sine^(4x))^(-1/2) d/dx(sine^(4x)) #

#=1/2(sine^(4x))^(-1/2)cose^(4x) . d/dx(e^(4x))#

#=1/2(sine^(4x))^(-1/2)cose^(4x). e^(4x) . d/dx(4x) #

#=1/2(sine^(4x))^(-1/2) .cose^(4x) . e^(4x) . 4 #

#=( 2.e^(4x).cose^(4x))/(sqrt(sine^(4x))#