How do you find the derivative of cot x?
The cot or cotangent function is the reciprocal of the tangent function. Mathematically, it can be represented as cot x = 1/tan x. Now, we know that tan x = sin x/cos x, so we can write cot x = cos x/sin x.
To find the derivative if cot x, we are using three different methods.
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Derivative using Chain Rule
Since cot x = 1/tan x, we write
Now we know that tan x = sin x/cos x and sec x = 1/cos x. So, the derivative turns out to be
This can be simplified as -1/sin2x = -cosec2x.
So, (d/dx)cot x = -cosec2x
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Derivative using Quotient Rule
We can write cot x = cos x/sin x
So, (d/dx)cot x = (d/dx)(cos x/sin x)
This is an u/v form. According to quotient rule, if u and v are functions of x, then
(d/dx)(u/v) = (u’v-v’u)/v2
Here, u = cos x and v = sin x. So, u’ = -sin x and v’ = cos x
So,
Since sin2 x + cos2 x = 1, we write the above expression as -1/sin2 x = -cosec2 x
So, (d/dx)cot x = -cosec2x
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Derivative Using First Principle
In this method, we use the basic definition of a derivative. For a function f(x), its derivative is written as
So, considering f(x) = cot x, we get
We know the addition subtraction formula in trigonometry, which states
sin (A-B) = sin A cos B – cos A sin B. Applying this formula in the above expression, we get
We can write this, since sin (-x) = -sin x
Using limits, we know
So, we write
So, (d/dx)cot x = -cosec2x
As we can see from these methods, the derivative of cot x is -cosec2x.