Thread: Find Derivative

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!

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: $\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}$

$\displaystyle \frac{(1+\cot x)(-\csc^2x)-(\cot x)(-\csc^2x)}{(1+\cot x)^2}=\frac{-\csc^2x(1+\cot x-\cot x)}{(1+\cot x)^2} = \frac{-\csc^2x}{(1+\cot x)^2}$