d/dx(cotx)
To find the derivative of cot(x), we can use the quotient rule
To find the derivative of cot(x), we can 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 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). To find the derivatives g'(x) and h'(x), we can use the chain rule.
The derivative of g(x) = 1 is simply 0 because it is a constant.
The derivative of h(x) = tan(x) can be found using the chain rule. Let u = x and v = tan(u). Then,
h'(x) = (d/dx) tan(x) = (d/dx) tan(u) = (d/dx) (v)
Using the chain rule, we have:
h'(x) = (dv/du) * (du/dx)
To find (dv/du), we differentiate tan(u) with respect to u:
(dv/du) = (d/du) tan(u) = sec^2(u)
To find (du/dx), we differentiate u = x with respect to x:
(du/dx) = (d/dx) x = 1
Now, we can substitute all these values into the quotient rule formula:
f'(x) = (g'(x)h(x) – g(x)h'(x)) / (h(x))^2
= (0*tan(x) – 1*sec^2(x)) / (tan(x))^2
= -sec^2(x) / tan^2(x)
We can simplify this expression further using trigonometric identities. Recall that sec^2(x) = 1 + tan^2(x). Substituting this into the expression:
f'(x) = -sec^2(x) / tan^2(x)
= -(1 + tan^2(x)) / tan^2(x)
= -1/tan^2(x) – 1
Finally, recall that cot(x) is the reciprocal of tan(x), so we can rewrite the derivative:
f'(x) = -1/(tan^2(x)) – 1
= -cot^2(x) – 1
Therefore, the derivative of cot(x) is -cot^2(x) – 1.
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