d/dx [cot x]=
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 in the form of f(x)/g(x), then the derivative will be (g(x)f'(x) – f(x)g'(x))/(g(x))^2.
In this case, we have f(x) = 1 and g(x) = tan x. We can rewrite cot x as 1/tan x, so f(x) = 1 and g(x) = tan x.
To find f'(x) and g'(x), we need to differentiate f(x) and g(x) with respect to x:
f'(x) = d/dx (1) = 0 (the derivative of a constant is zero)
g'(x) = d/dx (tan x)
To differentiate tan x, we can use the chain rule. Let u = x, and let y = tan u. Then, dy/du = sec^2 u, and du/dx = 1.
Using the chain rule, we have dy/dx = (dy/du)(du/dx) = sec^2 u * 1 = sec^2 x.
Therefore, g'(x) = d/dx (tan x) = sec^2 x.
Now we have all the parts we need to apply the quotient rule:
d/dx (cot x) = (g(x)f'(x) – f(x)g'(x))/(g(x))^2
= (tan x * 0 – 1 * sec^2 x)/(tan x)^2
= -sec^2 x / (tan^2 x)
= -1/cos^2 x.
Therefore, the derivative of cot x with respect to x is -1/cos^2 x.
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