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Mathematics 20 Online
OpenStudy (anonymous):

Find the derivative of y= cot x/ (1+cot x)

OpenStudy (anonymous):

quotient rule for this one

OpenStudy (anonymous):

Ohhh. Right, I forgot about that rule. Its v'*u-v*u'/ v^2 ?

OpenStudy (anonymous):

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with \[f(x)=\cot(x), f'(x)=-\csc^2(x),g(x)=1+\cot(x),g'(x)=-\csc^2(x)\]

OpenStudy (anonymous):

thanks!

OpenStudy (anonymous):

yw

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