Question

How do you find the derivative of `(ln tan(x))^2`?

Answer 1

How? Use the chain rule -- twice.
And simplify the answer. (Often simplifying is the tricky part.)

The derivative of some stuff squared is two times the stuff, times the derivative of that stuff.
Formally : the derivative of `[g(x)]^2` is `2[g(x)]*g'(x)`

In this case the "inside stuff", the `g(x)` is itself, the `ln` of some stuff. So to find the derivative, we'll need the chain rule again.
The derivative of the `la` of some stuff is `1` over the stuff, times the derivative of that stuff.
Formally : the derivative of `ln[f(x)]` is `1/f(x)*f'(x)`

Finally, we need to know the derivative of `tanx`.

The derivative of `(ln tan(x))^2` is

`2(ln tan(x))*1/tan(x)*sec^2(x)`

Now, we've finished the calculus, but we can re-write our answer in simpler form.
`1/tan(x) = cot(x)`
and `cot(x) = cos(x)/sin(x)`.

The second equality wouldn't be helpful, except that `sec(x)=1/cos(x)`.

Therefore we can re-write the derivative using:

`1/tan(x)*sec^2(x) = cot(x)sec^2(x)=cos(x)/sin(x)*1/cos^2(x)=1/sin(x)*1/cos(x)`

Which is surely more simply written as `csc(x)sec(x)`

So
The derivative of `(ln tan(x))^2` is

`2(ln tan(x))csc(x)sec(x)`

(If there's any advantage to it, we could also re-write `ln tan(x)` as `ln(sinx)-ln(cosx)`.)