d/dx(cotx)
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 we have a function of the form `f(x) = g(x) / h(x)`, then the derivative of `f(x)` is given by:
f'(x) = (g'(x) * h(x) – g(x) * h'(x)) / (h(x))^2
In this case, `g(x) = 1` and `h(x) = tan(x)`, so we have:
cot(x) = 1 / tan(x)
Now we will differentiate both `g(x)` and `h(x)`:
g'(x) = 0 (the derivative of a constant is zero)
h'(x) = sec^2(x) (the derivative of tan(x) is sec^2(x))
Now we can substitute the values into the quotient rule:
[f'(x) = (g'(x) * h(x) – g(x) * h'(x)) / (h(x))^2]
[f'(x) = (0 * tan(x) – 1 * sec^2(x)) / (tan(x))^2]
Simplifying further:
[f'(x) = -sec^2(x) / tan^2(x)]
Since `cot(x) = 1 / tan(x)`, we can simplify the expression to:
[f'(x) = -sec^2(x) / (1 / cot^2(x))]
Multiplying by the reciprocal:
[f'(x) = -sec^2(x) * cot^2(x)]
Therefore, the derivative of `cot(x)` is `-sec^2(x) * cot^2(x)`.
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