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

1 Answer
Feb 8, 2016

Answer:

#f'(x) =( 2e^(4x)sec^2e^(4x))/sqrt(tane^(4x)#

Explanation:

making use of the chain rule

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

and knowing # d/dx(tanx) = sec^2xcolor(black)(" and ") d/dx(e^x) = e^x#

first step is to rewrite #sqrt(tane^(4x)) = (tane^(4x))^(1/2) #

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

now # d/dx(tane^(4x)) = sec^2e^(4x) d/dx(e^(4x)).............(1)#

and #d/dx(e^(4x)) = e^(4x) d/dx(4x) = 4e^(4x)................(2) #

replacing the results of (1) and (2) back into f'(x)

#f'(x) = 1/2(tane^(4x))^(-1/2) . sec^2e^(4x) . 4e^(4x) #

and rewriting negative exponent as radical

# rArr f'(x) =( 2e^(4x)sec^2e^(4x))/sqrt(tane^(4x) #