How do you find the derivative of $y = cosx(cotx)$ ?

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Answer 1 · 2016-07-20T07:26:51

$(dy)/(dx) = cosx(-csc^2x-1)$

We have to use the product rule.

$d/(dx)(uv) = u(dv)/(dx) + (du)/(dx)v$

$(dy)/(dx) = cosx(d/(dx)(cotx)) + d/(dx)(cosx)cotx$

$= -cosxcsc^2x - sinxcotx$

Remember that $cotx = 1/tanx = cosx/sinx$

$therefore (dy)/(dx) = =-cosxcsc^2x - cosx$

How do you find the derivative of y = cosx(cotx) y = cosx(cotx)?