find derivative of the function (do not simplify the answer) please?

Question

f(x)=ln^2(1+cos^2(5x))

Answer 1

Some care is needed in dealing with this brute of a function, clearly designed to get your namesake's knickers in a twist, as we say in the old country.

You say don't simplify; I won't, then. I'll simply proceed "from the outside in," as we gradually unravel this Russian doll of successive functional dependencies, using the chain rule for differentiation.

Since it's questioned above, I'm assuming that the function can also be written as:

F(x) = [ln (1+cos^2(5x))]^2; O.K.?

Then the derivative is: df/dx =

2 ln (1+cos^2(5x)) * 1 / (1+cos^2(5x)) * (- 2 sin(5x) cos (5x) ) * 5.

In this expression, the *'s separate out the successive intermediate differentiations being performed as one peels away the successive functional forms, differentiating each new "outer function" as one goes at them, via the chain rule.

Phew! That really WAS an awkward function to deal with! (I've not done anything like this for over 40 years, since my graduate days at Starfleet Academy.)

Live long and prosper.

Answer 2

I'm having difficulty understanding what the ^2 means. I'm gonna take it to mean that the ^2 means all of ln squared.

f(x) = [ln(1 + cos^2(5x))]^2

Therefore, using a combination of the power rule and chain rule

f'(x) = 2[ln(1 + cos^2(5x))] [1/(1 + cos^2(5x))] [2cos(5x)][-sin(5x)][5]

Answer 3

Rewrite as y = 9t^(-a million/2) then you truly can word the skill rule. Multiply with the help of the exponent (-a million/2), then cut back it with the help of a million from -a million/2 to -3/2. y' = -9/2 t^-3/2 In radical variety, this could be y' = -9?t/2t^2

Answer 4

f'(x) = {1/(2(1+cos^2(5x))}*{2(1-2cos5xsin5x)}