Derivative of cot(x)
To find the derivative of cot(x), we will use the quotient rule
To find the derivative of cot(x), we will use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by:
f'(x) = (g'(x) * h(x) – h'(x) * g(x)) / (h(x))^2
In this case, g(x) = 1 and h(x) = tan(x).
Let’s find the derivatives of g(x) and h(x):
g'(x) = 0, because 1 is a constant function and its derivative is always zero.
h'(x) = sec^2(x), since the derivative of tan(x) is sec^2(x) (a well-known trigonometric derivative).
Now we can use the quotient rule to find the derivative of cot(x):
f'(x) = (g'(x) * h(x) – h'(x) * g(x)) / (h(x))^2
= (0 * tan(x) – sec^2(x) * 1) / (tan(x))^2
= -sec^2(x) / tan^2(x)
Now, we need to simplify this expression using trigonometric identities.
Recall that sec^2(x) = 1 + tan^2(x) and tan^2(x) = sec^2(x) – 1.
Substituting these identities into the derivative expression, we get:
f'(x) = -sec^2(x) / tan^2(x)
= -(1 + tan^2(x)) / (sec^2(x) – 1)
= -1
Therefore, the derivative of cot(x) is -1.
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