d/dx(cotx)
To find the derivative of the function f(x) = cot(x), where cot(x) represents the cotangent of x, we can use the quotient rule
To find the derivative of the function f(x) = cot(x), where cot(x) represents the cotangent of x, we can use the quotient rule. The quotient rule states that for a function of the form f(x) = g(x)/h(x), the derivative is given by:
f'(x) = (g'(x)h(x) – g(x)h'(x))/[h(x)]^2
To apply the quotient rule, we need to find the derivatives of g(x) and h(x). In this case, g(x) = 1 (since cot(x) can be written as 1/tan(x)), and h(x) = tan(x).
First, we find g'(x) = 0 (since the derivative of a constant is zero).
Next, we find h'(x) using the chain rule. The chain rule states that if we have a composition of two functions, f(g(x)), then the derivative is given by:
(f(g(x)))’ = f'(g(x)) * g'(x)
In this case, g(x) = tan(x) and f(x) = h(x) = x, so applying the chain rule:
h'(x) = 1 * sec^2(x) = sec^2(x)
Now, we can substitute these values into the quotient rule formula:
f'(x) = (0 * tan(x) – 1 * sec^2(x))/[tan(x)]^2
Simplifying further:
f'(x) = -[sec^2(x)]/[tan^2(x)]
To simplify this further, we can use the identity that sec^2(x) = 1 + tan^2(x):
f'(x) = -(1 + tan^2(x))/[tan^2(x)]
Multiplying both the numerator and denominator by -1, we get:
f'(x) = (tan^2(x) – 1)/[tan^2(x)]
Further simplifying, we have:
f'(x) = (tan(x))^2/[tan^2(x)]
Finally, we can cancel out the common factor of tan^2(x):
f'(x) = 1
So, the derivative of cot(x) with respect to x is simply 1.
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