Question

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

Answer 1

`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)`