Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 30

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


$
\begin{equation}
\begin{aligned}

\text{if } y =& (\ln \tan x), \text{ then by applying Chain Rule}
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y' =& 2 (\ln \tan x) \cdot \frac{d}{dx} (\ln \tan x)
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y' =& (\ln \tan x) \cdot \left( \frac{\displaystyle \frac{d}{dx} (\tan x)}{\tan x} \right)
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y' = & 2 (\ln \tan x) \left( \frac{\sec^2x}{\tan x} \right)
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y' =& 2 (\ln \tan x) \left( \frac{\displaystyle \frac{1}{\cos^2 x}}{\displaystyle \frac{\sin x}{\cos x}} \right)
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y' =& 2 (\ln \tan x) \left( \frac{1}{\cos ^ 2 x} \cdot \frac{\cos x }{\sin x} \right)
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y' =& 2 (\ln \tan x) \left( \frac{1}{\sin x \cos x} \right)
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y' =& 2 \csc x \sec x (\ln \tan x)

\end{aligned}
\end{equation}
$