Question

How do you differentiate `f(x)=cot(ln2x)` using the chain rule?

Answer 1

`f'(x) = -csc^2(ln(2x))/x`

Explanation:

Substitute `u = ln(2x) = ln(2) + ln(x)`.

`frac{"d"u}{"d"x} = 1/x`

Now substitute `x` and differentiate.

`f'(x) = frac{"d"}{"d"x}(cot(ln(2x)))`

`= frac{"d"}{"d"x}(cot(u))`

`= frac{"d"}{"d"u}(cot(u)) frac{"d"u}{"d"x}`

`= -csc^2(u) 1/x`

`= -csc^2(ln(2x))/x`