## derivative of cotx

### -csc^2x

The derivative of cot(x) can be found using the quotient rule of differentiation.

Let y = cot(x)

Using the identity, cot(x) = cos(x)/sin(x), we can rewrite y as:

y = cos(x)/sin(x)

Now, using the quotient rule of differentiation, we get:

y’ = [(cos(x))(sin(x))’ – (sin(x))(cos(x))’] / (sin(x))^2

Taking the derivatives of sin(x) and cos(x), we get:

sin(x)’ = cos(x) and cos(x)’ = -sin(x)

Substituting these values in the above equation, we get:

y’ = [(cos(x))(cos(x)) – (sin(x))(-sin(x))] / (sin(x))^2

Simplifying the equation, we get:

y’ = [cos^2(x) + sin^2(x)] / (sin(x))^2

Using the identity, cos^2(x) + sin^2(x) = 1, we can further simplify the equation as:

y’ = 1 / (sin(x))^2

Therefore, the derivative of cot(x) is:

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

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