Question

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

Answer 1

`-2"cosec"^2 x. cotx`

Explanation:

`f(x) = cot^2x`

`d [f(x)]/dx = 2cot x.(-"cosec"^2x).1`

`= -2"cosec"^2 x. cotx`

first you get the derivative of the indice, then the trig function, lastly the x.
it's easy if you remember it that way ;)

Answer 2

`-(2cotx)/sin^2x`

Explanation:

I like the previous answer, but I think it's simpler to use the product rule. Since `cot^2x=cotx*cotx`, `d/dx(cot^2x)=f'(x)g(x)+g'(x)f(x)`, where `f(x)=g(x)=cotx`. This evaluates to `cotx xx (-1/sin^2x)+(-1/sin^2x)xxcotx =2(-1/sin^2x)cotx`, `-(2cotx)/sin^2x`. This is equivalent to the previous answer, since `1/sin^2x="cosec" ^2x`, but I think that that answer is cleaner, don't you?