Thread: derivative of cot x
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Old November 7th, 2009, 03:23 PM
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Quote:
Originally Posted by saweetness View Post
use the quotient rule to show that
d/dx cotx = -csc^2 x
(hint: Write cotx = cosx/sinx)

...

cotx = cosx/sinx

:S

please help THANKS!
The quotient rule says that:
\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{[g(x)]^2}

f(x) = cos(x)
g(x) = sin(x)
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