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

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Answer 1 · 2016-07-20T07:26:51

$\frac{d y}{d x} = cos x (- {csc}^{2} x - 1)$ $(dy)/(dx) = cosx(-csc^2x-1)$

We have to use the product rule.

$\frac{d}{d x} (u v) = u \frac{d v}{d x} + \frac{d u}{d x} v$ $d/(dx)(uv) = u(dv)/(dx) + (du)/(dx)v$

$\frac{d y}{d x} = cos x (\frac{d}{d x} (cot x)) + \frac{d}{d x} (cos x) cot x$ $(dy)/(dx) = cosx(d/(dx)(cotx)) + d/(dx)(cosx)cotx$

$= - cos x {csc}^{2} x - sin x cot x$ $= -cosxcsc^2x - sinxcotx$

Remember that $cot x = \frac{1}{tan x} = \frac{cos x}{sin x}$ $cotx = 1/tanx = cosx/sinx$

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

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