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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