Derivative of cot(x)
-csc^2u du
The derivative of cot(x) can be found using the quotient rule. Recall that:
(1) The quotient rule states that if f(x) and g(x) are differentiable functions, then the derivative of f(x)/g(x) is given by [(g(x)*f'(x)) – (f(x)*g'(x))]/[g(x)]^2.
Now, let’s apply the quotient rule to find the derivative of cot(x):
Let f(x) = 1 and g(x) = tan(x). Then, using the trigonometric identity cot(x) = cos(x)/sin(x), we can write cot(x) as:
cot(x) = cos(x)/sin(x) = f(x)/g(x)
Now, we need to find f'(x) and g'(x):
f'(x) = 0, since the derivative of a constant is 0.
g'(x) = sec^2(x), since the derivative of tan(x) is sec^2(x).
Using the quotient rule with these values, we get:
cot'(x) = [(sin(x)*0) – (cos(x)*sec^2(x))]/[sin(x)]^2
Simplifying this expression, we get:
cot'(x) = -cos(x)*csc^2(x)
Therefore, the derivative of cot(x) is -cos(x)*csc^2(x).
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