derivative of cotx
-csc^2x
The derivative of cot x can be found using the quotient rule. Let us denote certain trigonometric identities:
cosec^2 x = 1/sin^2 x,
cot x = cos x / sin x
Then, we can write:
d/dx[cot x] = d/dx[cos x / sin x]
Using the quotient rule, we get:
= [(-sin x)* d/dx[cos x] – (cos x)* d/dx[sin x] ] / (sin x)^2
Now, we need to find the derivative of cos x and sin x separately.
d/dx[cos x] = -sin x
d/dx[sin x] = cos x
Putting these values, we get:
= [(-sin x)*(-sin x) – (cos x)*cos x] / (sin x)^2
= [sin^2 x + cos^2 x] / (sin x)^2
= (1/sin^2 x)
Therefore,
d/dx[cot x] = -cosec^2 x = -1/(sin^2 x)
Hence, the derivative of cot x is -cosec^2 x or -1/(sin^2 x).
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