- #1
ultima9999
- 43
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Yeah, I was working through this problem and it differs from the answer that my friend got.
Using implicit differentiation, find the derivative of [tex]\mbox{arc}\tanh \frac{x}{2}[/tex] and state the domain for which the derivative applies
[tex]\begin{align*}
y = \arctanh \frac{x}{2} \\
\Leftrightarrow x = 2 \tanh y
\end{align*}[/tex]
[tex]\frac{d}{dx}x = \frac{d}{dx}2 \tanh y[/tex]
[tex]\Rightarrow 1 = 2\ \mbox{sech}^2 y \cdot \frac{dy}{dx}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2\ \mbox{sech}^2 y}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2 - 2 \tanh^2 y}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2 - 2 \frac{x^2}{4}}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2 - \frac{x^2}{2}}[/tex]
Using implicit differentiation, find the derivative of [tex]\mbox{arc}\tanh \frac{x}{2}[/tex] and state the domain for which the derivative applies
[tex]\begin{align*}
y = \arctanh \frac{x}{2} \\
\Leftrightarrow x = 2 \tanh y
\end{align*}[/tex]
[tex]\frac{d}{dx}x = \frac{d}{dx}2 \tanh y[/tex]
[tex]\Rightarrow 1 = 2\ \mbox{sech}^2 y \cdot \frac{dy}{dx}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2\ \mbox{sech}^2 y}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2 - 2 \tanh^2 y}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2 - 2 \frac{x^2}{4}}[/tex]
[tex]\Rightarrow \frac{dy}{dx} = \frac{1}{2 - \frac{x^2}{2}}[/tex]
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