What is the derivative of $y = ln ({sin}^{2} (θ))$ $y = ln(sin^2(theta))$ ?

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Answer 1 · 2016-10-09T21:43:47

${cot}^{2} θ$ $Cot^2theta$

Using the chain rule we have:

$F(x)'= f'(g(x))*g'(x)$

$y'=ln'(sin^2(theta))*(cos^2(theta))$
$1/sin^2theta *Cos^2theta$

$Cot^2theta$

What is the derivative of y = ln(sin^2(theta))y=ln(sin2(θ))y = ln(sin^2(theta))?