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 -csc^{2}x.
Find the derivative of each of the following.
Problem 1 :
cot(2x)
Solution :
We already know the derivative of cotx, which is -csc^{2}x. We can find the derivative of cot(2x) using chain rule.
= [cot(2x)]'
= [-csc^{2}(2x)](2x)'
= [-csc^{2}(2x)](2)
= -2csc^{2}(2x)
Problem 2 :
cot(3x + 5)
Solution :
= [cot(3x + 5)]'
= [-csc^{2}(3x + 5)](3x + 5)'
= [-csc^{2}(3x + 5)](3)
= -3csc^{2}(3x + 5)
Problem 3 :
cot(x^{2} + 3x + 5)
Solution :
= [cot(x^{2} + 3x + 5)]'
= [-csc^{2}(x^{2} + 3x + 5)](x^{2} + 3x + 5)'
= [-csc^{2}(x^{2} + 3x + 5)](2x + 3)
= -(2x + 3)csc^{2}(x^{2} + 3x + 5)
Problem 4 :
cot^{2}x
Solution :
= (cot^{2}x)'
= (2cot^{2-1}x)(cotx)'
= (2cotx)(-csc^{2}x)
= -2(cotx)(csc^{2}x)
Problem 5 :
Solution :
Problem 6 :
cot√x
Solution :
Problem 7 :
e^{cotx}
Solution :
= (e^{cotx})'
= e^{cotx}(cotx)'
= e^{cotx}(-csc^{2}x)
= -(csc^{2}x)e^{cotx}
Problem 8 :
ln(cotx)
Solution :
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