Question

Question #056b8

Answer 1

Apply the chain rule twice to find `f'(x) = (20ln^3(5x-1))/(5x-1)`

Explanation:

To solve this, we will use the chain rule, which states
`d/dx f(g(x)) = f'(g(x))g'(x)`
Let
`f_1(x) = x^4`
`f_2(x) = ln(x)`
`f_3(x) = 5x - 1`
`f_4(x) = ln(5x-1) = f_2(f_3(x))`

Now, applying the chain rule,
`d/dx ln^4(5x-1) = d/dx f_1(f_4(x)) = f_1'(f_4(x))f_4'(x)`

`f_1'(x) = 4x^3` so `f_1'(f_4(x)) = 4ln^3(5x-1)`

To find `f_4'(x)` we once again use the chain rule.

`f_4'(x) = d/dx f_2(f_3(x)) = f_2'(f_3(x))f_3'(x)`

`f_2'(x) = 1/x` and `f_3'(x) = 5`

So we have
`f_4'(x) = 1/(5x-1)*5= 5/(5x-1)`

Substituting back into the original equation gives us the final result

`d/dx ln^4(5x-1) = 4ln^3(5x-1)*5/(5x-1)=(20ln^3(5x-1))/(5x-1)`