Single Variable Calculus

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Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 32

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Differentiate $\displaystyle y = \ln \tan ^2 x $

$ \begin{equation} \begin{aligned} \text{if } y =& \ln \tan ^2 x, \text{ then } \\ \\ y' =& \frac{\displaystyle \frac{d}{dx} (\tan^2 x) }{\tan^2 x} \\ \\ y' =& \frac{2 (\tan x)(\sec^2 x)}{\tan ^2 x} \\ \\ y' =& \frac{2 \sec^2 x}{\tan x} \\ \\ y' =& \frac{\displaystyle 2 \left( \frac{1}{\cos ^ 2x} \right)}{\displaystyle \frac{\sin x}{\cos x}} \end{aligned} \end{equation} $

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