Question

What is the derivative of `6/tan(2x)`?

Answer 1

`f'(x)=-12cosec^2(2x)`

Explanation:

Given that we will need to know a few results here to solve the given question. The results will be
`\frac{d}{dx}(x^n)=nx^(n-1)`
`frac{d}{dx}(tan(kx))=k*sec^2(kx)`

So, given `y=6/tan(2x)`
So, `\frac{d}{dx}(y)=(-6)/tan^2(2x)*2sec^2(2x)`

We also know that `tanx=sinx/cosx` and that `secx=1/cosx`
So, we expand the terms such that
`f'(x)=(-6)/(sin(2x)/cos(2x))^2*2(1/cos^2(2x))`
So by cancelling the `cos(2x)` function and rearranging, we get the above given answer.