Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 40

Determine the derivative of $\displaystyle y = \sqrt[4]{\frac{1 + \tan hx}{1 - \tan hx}}$. Simplify where possible.


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\begin{equation}
\begin{aligned}

y' =& \frac{d}{dx} \left( \sqrt[4]{\frac{1 + \tan hx}{1 - \tan hx}} \right)
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y' =& \frac{d}{dx} \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{1}{4}}
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y' =& \frac{1}{4} \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{-3}{4}} \cdot \frac{d}{dx} \left( \frac{1 + \tan hx}{1 - \tan hx} \right)
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y' =& \frac{1}{4} \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{-3}{4}} \left[ \frac{\displaystyle (1 - \tan hx) \frac{d}{dx} (1 + \tan hx) - (1 + \tan hx) \frac{d}{dx} (1 - \tan hx) }{(1 - \tan hx)^2} \right]
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y' =& \frac{1}{4} \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{-3}{4}} \left[ \frac{(1 - \tan hx)(\sec h^2 x) - (1 + \tan hx)(- \sec h^2 x)}{(1 - \tan hx)^2} \right]
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y' =& \frac{1}{4} \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{-3}{4}} \left[ \frac{\sec h^2 x - \cancel{\sec h^2 x \tan hx} + \sec h^2 x + \cancel{\sec h^2 x \tan hx}}{(1 - \tan hx)^2} \right]
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y' =& \frac{1}{4} \cdot \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{-3}{4}} \left[ \frac{2 \sec h^2 x}{(1 - \tan hx)^2} \right]
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y' =& \frac{1}{4} \cdot \frac{1}{\displaystyle \left( \frac{1 + \tan hx}{1 - \tan hx} \right)^{\frac{3}{4}}} \left[ \frac{2 \sec h^2 x}{(1 - \tan hx)^2} \right]
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y' =& \frac{1}{2} \cdot \frac{1}{\displaystyle \frac{(1 + \tan hx)^{\frac{3}{4}}}{(1 - \tan hx)^{\frac{3}{4}}}} \cdot \frac{\sec h^2 x}{(1 - \tan hx)^2}
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y' =& \frac{(1 - \tan hx)^{\frac{3}{4}}}{2 (1 + \tan hx)^{\frac{3}{4}}} \cdot \frac{\sec h^2 x}{(1 - \tan hx)^2}
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y' =& \frac{\sec h^2 x}{2 (1 + \tan hx)^{\frac{3}{4}} (1 - \tan hx)^{\frac{5}{4}}}
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y' =& \frac{\sec h^2 x}{2 (1 + \tan hx)^{\frac{3}{4}} (1 - \tan hx)^{\frac{3}{4}} (1 - \tan hx)^{\frac{2}{4}}}
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y' =& \frac{\sec h^2 x}{2 [(1 + \tan hx)(1 - \tan hx)]^{\frac{3}{4}} (1 - \tan hx)^{\frac{1}{2}} }
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y' =& \frac{\sec h^2 x}{2 (1 - \tan h^2 x)^{\frac{3}{4}} (1 - \tan hx)^{\frac{1}{2}}}
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y' =& \frac{\sec h^2 x}{2 (\sec h^2 x)^{\frac{3}{4}} (1 - \tan hx)^{\frac{1}{2}}}
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y' =& \frac{(\sec h^2 x)^{\frac{1}{4}}}{2 (1 - \tan hx)^{\frac{1}{2}}}
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y' =& \frac{(\sec hx)^{\frac{2}{4}}}{2 (1 - \tan hx)^{\frac{1}{2}}}
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y' =& \frac{(\sec hx)^{\frac{1}{2}}}{2 (1 - \tan hx)^{\frac{1}{2}}}
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& \text{or}
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y' =& \frac{1}{2} \sqrt{\frac{\sec hx}{1 - \tan hx}}

\end{aligned}
\end{equation}
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