Question

y=cot(x) what does y'=

Answer 1

do you remember the derivative of tangent?

Answer 2

you find the derivative of a "co function" by changing to "co" and using a minus sign. so if you know that the derivative of tangent is secant squared, then the derivative of cotangent is
minus cosecant squared

Answer 3

I know but I am supposed to prove so what are the steps to find it?

Answer 4

what are you allowed to use?

Answer 5

just prove that cot'(x) = -csc^2(x) using something like the quotient rule

Answer 6

i would use this
\[\cot(x)=\frac{1}{\tan(x)}\] so
\[\cot'(x)=\frac{-1}{\tan^2(x)}\times \sec^2(x)\] by the rule for reciprocals, the quotient rule or the chain rule whichever one you care to cite

Answer 7

then simplify this mess to get
\[-\csc^2(x)\]

Answer 8

why did you say the cot'(x) = -1/(tan^2(x))*sec^2(x)

Answer 9

you have a choice of reasons. if you didn't get to the chain rule yet you can use the quotient rule

Answer 10

\[(\frac{1}{f})'=-\frac{f'(x)}{f^2(x)}\]

Answer 11

ok I understand it thank you