The cotangent of x is defined as cosine of x divided by sine of x.
cotx = ᶜᵒˢˣ⁄sᵢnₓ
Since cotx can be written as cosx divided by sinx, we can find the derivative of cotx using quotient rule.
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.
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 :
cot√x
Solution :
Problem 7 :
ecotx
Solution :
= (ecotx)'
= ecotx(cotx)'
= ecotx(-csc2x)
= -(csc2x)ecotx
Problem 8 :
ln(cotx)
Solution :
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