Differentiating cotx

  • #1
33
1
I keep getting the wrong answer when i try to differentiate cotx..
this is what i get:
cotx = 1/tanx =cosx/sinx=cosx ⋅ sin^-1
so by the product and chain rule we have:
sinx⋅(sin x)^-1+cos⋅(-1sin^2 x)^-1 ⋅(cosx)^-1

=

sinx/sinx - cosx/cosx ⋅ sin^2x
=1-1/sin^2 x

where as the correct answer is -1/sin^2x = -csc^2 x

could someone please tell me where i am going wrong?


many thanks
Ryan
 
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Answers and Replies

  • #2
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I keep getting the wrong answer when i try to differentiate cotx..
this is what i get:
cotx = 1/tanx =cosx/sinx=cosx ⋅ sin^-1
so by the product and chain rule we have:
sinx⋅(sin x)^-1+cos⋅(-1sin^2 x)^-1 ⋅(cosx)^-1
d/dx(cos(x)) = -sin(x). It looks like you have other mistakes as well. For this problem it's probably simpler to use the quotient rule. You don't need to use the chain rule when you do so.
wolfspirit said:
=

sinx/sinx - cosx/cosx ⋅ sin^2x
=1-1/sin^2 x

where as the correct answer is -1/sin^2x = -csc^2 x

could someone please tell me where i am going wrong?


many thanks
Ryan
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
I keep getting the wrong answer when i try to differentiate cotx..
this is what i get:
cotx = 1/tanx =cosx/sinx=cosx ⋅ sin^-1
so by the product and chain rule we have:
sinx⋅(sin x)^-1+cos⋅(-1sin^2 x)^-1 ⋅(cosx)^-1

=

sinx/sinx - cosx/cosx ⋅ sin^2x
=1-1/sin^2 x

where as the correct answer is -1/sin^2x = -csc^2 x

could someone please tell me where i am going wrong?


many thanks
Ryan

You have cot(x) = cos(x) * sin-1(x) = u * v

u = cos (x)
v = sin-1(x)

u' = -sin(x)
v' = -sin-2(x) * cos (x) [from the chain rule]

d(cot(x))/dx = u * v' + v * u' = -cos2(x)*sin-2(x) - sin(x) * sin-1(x) = -cot2(x) - 1 = -[1 + cot2(x)]

cot2(x) = cos2(x) / sin2(x)

1 + cot2(x) = 1 + cos2(x)/sin2(x) = [sin2(x) + cos2(x)] / sin2(x) = 1/sin2(x) = csc2(x)

-[1 + cot2(x)] = -csc2(x) = d(cot(x))/dx

Q.E.D.
 

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