d/dx [cot x]
To find the derivative of cot x, we can start by expressing it in terms of sine and cosine:
cot x = cos x / sin x
Now, let’s differentiate both sides with respect to x using the quotient rule:
(d/dx) [cot x] = (d/dx) [cos x / sin x]
Using the quotient rule, the derivative of the numerator, cos x, is -sin x, and the derivative of the denominator, sin x, is cos x
To find the derivative of cot x, we can start by expressing it in terms of sine and cosine:
cot x = cos x / sin x
Now, let’s differentiate both sides with respect to x using the quotient rule:
(d/dx) [cot x] = (d/dx) [cos x / sin x]
Using the quotient rule, the derivative of the numerator, cos x, is -sin x, and the derivative of the denominator, sin x, is cos x. Applying the quotient rule:
(d/dx) [cot x] = [(-sin x)(sin x) – (cos x)(cos x)] / (sin x)²
Simplifying the numerator:
(d/dx) [cot x] = [-sin² x – cos² x] / (sin x)²
Recall the Pythagorean identity, which states that sin² x + cos² x = 1. Rearranging this equation:
cos² x = 1 – sin² x
Substituting this identity into the numerator:
(d/dx) [cot x] = [-(1 – sin² x) – cos² x] / (sin x)²
(d/dx) [cot x] = [-1 + sin² x – cos² x] / (sin x)²
Since sin² x + cos² x = 1, we have:
(d/dx) [cot x] = [-1 + 1] / (sin x)²
Simplifying further:
(d/dx) [cot x] = 0 / (sin x)²
The final result is:
(d/dx) [cot x] = 0
Therefore, the derivative of cot x with respect to x is 0.
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