How do you differentiate #f(x)=sqrt(sine^(4x)# using the chain rule.?
1 Answer
Mar 8, 2016
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))#