d/dx cot x
-csc^2 x
Let us use the quotient rule of differentiation to find the derivative of the cotangent of x with respect to x.
Recall that the quotient rule states that for u = f(x) and v = g(x), the derivative of u/v with respect to x is:
(d/dx) [u(x)/v(x)] = [v(x)*(du/dx) – u(x)*(dv/dx)] / [v(x)]^2
In this case, we can let u = 1 (since cot x = 1/tan x) and v = tan x. Therefore:
(d/dx) [cot x] = (d/dx) [1/tan x]
= [-1*(d/dx) [tan x]] / [tan x]^2
Next, we need to find the derivative of tangent of x with respect to x, which is:
(d/dx) [tan x] = sec^2x
Substituting this back into our earlier equation, we get:
(d/dx) [cot x] = [-1*sec^2x] / [tan x]^2
Now, using the trigonometric identity tan^2x + 1 = sec^2x, we can simplify this expression as follows:
(d/dx) [cot x] = [-1*(1 + tan^2x)] / [tan x]^2
= -[1/tan^2x + 1]
Therefore, the derivative of cotangent of x with respect to x is:
(d/dx) [cot x] = -cosec^2x.
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