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

1 Answer
Jan 22, 2016

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.