derivative of cotx
cot'(x) = -sin(x) / cos^2(x)
The derivative of cot(x) can be found using the quotient rule of differentiation. Recall that the quotient rule states that if we have two functions f(x) and g(x), the derivative of their quotient (i.e., f(x)/g(x)) is given by:
(f'(x)g(x) – g'(x)f(x)) / g(x)^2
Let’s apply this rule to find the derivative of cot(x). First, we need to express cot(x) in terms of sine and cosine:
cot(x) = cos(x)/sin(x)
Now we can use the quotient rule to find the derivative:
cot'(x) = [(cos'(x)sin(x) – sin'(x)cos(x)) / sin(x)^2]
Next, we need to find the derivatives of cos(x) and sin(x). Recall that the derivative of cosine is negative sine, and the derivative of sine is cosine:
cos'(x) = -sin(x)
sin'(x) = cos(x)
Substituting these values, we get:
cot'(x) = [(-sin(x)sin(x) – cos(x)cos(x)) / sin(x)^2]
Simplifying this expression, we get:
cot'(x) = (-sin^2(x) – cos^2(x)) / sin^2(x)
Recall that sin^2(x) + cos^2(x) = 1. Substituting this value, we get:
cot'(x) = -1 / sin^2(x)
Finally, we can express the answer in terms of cot(x) by substituting cot(x) = cos(x)/sin(x):
cot'(x) = -sin(x) / cos^2(x)
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