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

2 Answers

-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 ;)

-(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?