y = (cot x) / (1 + cot x) I got down to (sec x tan x) / (1 + cot x) ^2 I dont know if this is right... The correct answer is... (-csc^2x) / (1 + cot x)^2 Can someone help me out? Thanks!
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Originally Posted by Morgan82 y = (cot x) / (1 + cot x) I got down to (sec x tan x) / (1 + cot x) ^2 I dont know if this is right... The correct answer is... (-csc^2x) / (1 + cot x)^2 Can someone help me out? Thanks! Use the quotient rule:
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About LinkBacks(-\csc^2x)-(\cot x)(-\csc^2x)}{(1+\cot x)^2}=\frac{-\csc^2x(1+\cot x-\cot x)}{(1+\cot x)^2} = \frac{-\csc^2x}{(1+\cot x)^2})
