How do you find the derivative of $cot x$ $cotx$ ?

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Answer 1 · 2016-11-14T04:43:21

$\frac{d y}{d x} = - {csc}^{2} x$ $dy/dx = -csc^2x$

$y = cot x$ $y = cotx$

$y = \frac{1}{tan x}$ $y = 1/tanx$

$y = \frac{1}{\frac{sin x}{cos x}}$ $y = 1/(sinx/cosx)$

$y = \frac{cos x}{sin x}$ $y = cosx/sinx$

Letting $y = \frac{g (x)}{h (x)}$ $y= (g(x))/(h(x))$ , we have that $g (x) = cos x$ $g(x) = cosx$ and $h (x) = sin x$ $h(x) = sinx$ .

$y' = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2$

$y' = (-sinx xx sinx - (cosx xx cosx))/(sinx)^2$

$y' = (-sin^2x - cos^2x)/(sinx)^2$

$y' = (-(sin^2x + cos^2x))/sin^2x$

$y' = -1/sin^2x$

$y' = -csc^2x$

Hopefully this helps!

How do you find the derivative of cotxcotxcotx?