Derivative of cot x
-csc^2 x
The derivative of cot x can be found using the quotient rule:
Let f(x) = cos x and g(x) = sin x
cot x can be written as f(x)/g(x)
Using the quotient rule, we have:
(cot x)’ = [f'(x) * g(x) – f(x) * g'(x)] / [g(x)]^2
f'(x) = -sin x and g'(x) = cos x
Plugging in the values, we get:
(cot x)’ = [(-sin x) * (sin x) – (cos x) * (cos x)] / [sin x]^2
Simplifying the equation, we have:
(cot x)’ = -[1 + (cos x / sin x)^2] / sin^2 x
Using trigonometric identity, cos^2 x + sin^2 x = 1, we have:
(cot x)’ = -[1 + cot^2 x] / sin^2 x
More Answers:
[next_post_link]