Prove that (cot x)' = - csc^2 x

We know that cosec x = 1/sin x and cot x = cos x/ sin
x

If we have to calculate the derivative of cot x, we'll
consider the quotient rule:

(cos x/sin x)' = (cos x)'*(sin
x) - (cos x)*(sin x)'/(sin x)^2

We'll calculate the
derivatives of the functions sine and cosine:

(cos x)' =
-sin x

(sin x)' = cos x

(cot
x)' = [(-sin x)*(sin x) - (cos x)^2]/(sin x)^2

(cot x)' = -
[(sin x)^2 + (cos x)^2]/(sin x)^2

But, from the fundamental
formula of trigonometry, we'll have:

[(sin x)^2 + (cos
x)^2] = 1

(cot x)' = -1/(sin
x)^2

But 1/sin x = csc x.

If
we'll raise to square both sides, we'll get:

1/(sin x)^2 =
-(csc x)^2