# Explain why d/dx cotx=-cscx^2?

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You need to prove that `(d(cot x))/dx = -csc^2 x` , hence you need to differentiate the cotangent function with respect to x such that:

`(cot x)' = (cos x/sin x)'`

Notice that you need to use the fact that the cotangent function is rational such that `cot x = cos x/sin x` .

You need to differentiate using quotient rule such that:

`(cot x)' = ((cos x)'*sin x -(cos x)*(sin x)')(sin^2 x)`

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

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

Using the basic trigonometric identity yields:

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

You need to remember that `1/sin x = csc x =gt 1/(sin^2 x)= csc^2 x` , hence `(cot x)' = -csc^2 x` .

**Hence, differentiating the cotangent function with respect to x yields: `(d(cot x))/dx = -csc^2 x.` **