# DERIVATIVE OF COTX USING QUOTIENT RULE

The cotangent of x is defined as cosine of x divided by sine of x.

cotx = ᶜᵒˢˣ⁄sn

Since cotx can be written as cosx divided by sinx, we can find the derivative of cotx using quotient rule.

## Quotient Rule of Derivative

Consider a function defined by y as shown below.

y = ᵘ⁄ᵥ

Let u and v be the functions of x.

Then the derivative of y with respect to x :

The above formula is called the Quotient Rule of Derivative.

Derivative of Cotx :

Substitute cotx for y, cosx for u and sinx for v into the above formula and and find the derivative of cotx.

Therefore, the derivative of cotangent of x is -csc2x.

## Solved Problems

Find the derivative of each of the following.

Problem 1 :

cot(2x)

Solution :

We already know the derivative of cotx, which is -csc2x. We can find the derivative of cot(2x) using chain rule.

= [cot(2x)]'

= [-csc2(2x)](2x)'

= [-csc2(2x)](2)

= -2csc2(2x)

Problem 2 :

cot(3x + 5)

Solution :

= [cot(3x + 5)]'

= [-csc2(3x + 5)](3x + 5)'

= [-csc2(3x + 5)](3)

= -3csc2(3x + 5)

Problem 3 :

cot(x2 + 3x + 5)

Solution :

= [cot(x2 + 3x + 5)]'

= [-csc2(x2 + 3x + 5)](x2 + 3x + 5)'

= [-csc2(x2 + 3x + 5)](2x + 3)

= -(2x + 3)csc2(x2 + 3x + 5)

Problem 4 :

cot2x

Solution :

= (cot2x)'

= (2cot2-1x)(cotx)'

= (2cotx)(-csc2x)

= -2(cotx)(csc2x)

Problem 5 :

Solution :

Problem 6 :

cotx

Solution :

Problem 7 :

ecotx

Solution :

= (ecotx)'

= ecotx(cotx)'

= ecotx(-csc2x)

= -(csc2x)ecotx

Problem 8 :

ln(cotx)

Solution :

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