Derivative of Cot(x) | Applying the Chain Rule and Quotient Rule in Calculus

d/dx(cotx)

To find the derivative of cot(x), we need to apply the chain rule

To find the derivative of cot(x), we need to apply the chain rule. Recall that the cotangent function can be written as the reciprocal of the tangent function, cot(x) = 1/tan(x).

Let’s start by using the quotient rule to find the derivative of tan(x). The tangent function is defined as tan(x) = sin(x) / cos(x).

Using the quotient rule, we have:

d/dx(tan(x)) = (cos(x) * d/dx(sin(x)) – sin(x) * d/dx(cos(x))) / (cos(x))^2.

Next, we need to find the derivatives of sin(x) and cos(x):

d/dx(sin(x)) = cos(x),

d/dx(cos(x)) = -sin(x).

Substituting these values into the derivative of tan(x), we get:

d/dx(tan(x)) = (cos(x) * cos(x) – sin(x) * (-sin(x))) / (cos(x))^2.

Simplifying the numerator, we have:

d/dx(tan(x)) = (cos^2(x) + sin^2(x)) / (cos(x))^2.

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can rewrite the numerator as:

d/dx(tan(x)) = 1 / (cos(x))^2.

Since cot(x) = 1/tan(x), we can substitute this into the equation to get:

d/dx(cot(x)) = d/dx(1/tan(x)) = -1 / (sin(x))^2.

Therefore, the derivative of cot(x) is -1 / (sin(x))^2.

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