Funciones hiperbólicas y sus inversas
En los problemas del 1 al 10, hallar la derivada \(\boldsymbol{y'= D_x y}\) de la función dada.
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\[ y = \tanh^{-1}(\cosh x) \]
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\[ y = \mathrm{e}^{ \text{ senh} (2x) } \]
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\[ y = x^{ \tanh x }, \; x > 0 \]
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\[ y = \frac{1}{2} \tanh \left( \frac{x}{2} \right) - \frac{1}{6} \tanh^{3} \left( \frac{x}{2}\right) \]
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\[ y = \mathrm{e}^{ ax } \cosh bx \]
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\[ y = \sqrt[4]{ \frac{ 1 + \tanh x }{ 1 - \tanh x } } \]
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\[ y = \left( \text{cosech}^{-1} x \right)^2 \]
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\[ y = \text{ senh}^{-1} \frac{x^2}{a^2} \]
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\[ y = \tanh^{-1} ( \sec x) \]
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\[ y = \tan^{-1} x + \tanh^{-1} x \]
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Probar las siguientes identidades dadas en el teorema 3.5.1:
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\[ \text{ senh } (-x) = - \text{ senh } x \]
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\[ \cosh (-x) = \cosh x \]
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\[ 1 - \coth^2 x = - \text{cosech}^2 x \]
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\[ \cosh (x + y) = \cosh x \cosh y + \text{ senh } x \text{ senh } y \]\[ \begin{aligned} \cosh (x + y) &= \cosh x \cosh y \\[.5em] & + \text{ senh } x \text{ senh } y \end{aligned} \]
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Probar las identidades:
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\[ \text{ senh } (x - y) = \text{ senh } x \cosh y - \cosh x \text{ senh } y \]\[ \begin{aligned} &\text{ senh } (x - y) \\[.5em] &\hspace{1em}= \text{ senh } x \cosh y \\[.5em] &\hspace{3em}- \cosh x \text{ senh } y \end{aligned} \]
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\[ \cosh (x - y) = \cosh x \cosh y - \text{ senh } x \text{ senh } y \]\[ \begin{aligned} &\cosh (x - y) \\[.5em] &\hspace{1em}= \cosh x \cosh y \\[.5em] &\hspace{3em}- \text{ senh } x \text{ senh } y \end{aligned} \]
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\[ \cosh x + \cosh y = 2\cosh \frac{x+y}{2} \cosh \frac{x-y}{2} \]\[ \begin{aligned} &\cosh x + \cosh y \\[.5em] &\hspace{1em} = 2\cosh \frac{x+y}{2} \cosh \frac{x-y}{2} \end{aligned} \]
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\[ \text{ senh } x + \text{ senh } y = 2 \text{ senh } \frac{x+y}{2} \cosh \frac{x-y}{2} \]\[ \begin{aligned} &\text{ senh } x + \text{ senh } y \\[.5em] &\hspace{1em}= 2 \text{ senh } \frac{x+y}{2} \cosh \frac{x-y}{2} \end{aligned} \]
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\[ \cosh x - \cosh y = 2 \text{ senh } \frac{x + y}{2} \text{ senh } \frac{x - y}{2} \]\[ \begin{aligned} &\cosh x - \cosh y \\[.5em] &\hspace{1em}= 2 \text{ senh } \frac{x + y}{2} \text{ senh } \frac{x - y}{2} \end{aligned} \]
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\[ \text{ senh } 3x = 3 \text{ senh } x + 4 \text{ senh }^3 x \]\[ \begin{aligned} &\text{ senh } 3x \\[.5em] &\hspace{1em}= 3 \text{ senh } x + 4 \text{ senh }^3 x \end{aligned} \]
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Probar las igualdades 2, 3, 4, 5 y 6 del teorema 3.5.4.
Ver Respuestas
Respuestas
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\[ y’ = -\text{cosech } x \]
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\[ y’ = 2 \cosh 2x \, \mathrm{e}^{\text{ senh } 2x} \]
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\[ y’ = x^{\tanh c} \left( \frac{\tanh x}{x} + \text{ sech}^2 x \ln x \right) \]\[ y’ = x^{\tanh c} \left( \frac{\tanh x}{x} + \text{ sech}^2 x \ln x \right) \]
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\[ y’ = \frac{1}{4} \text{ sech}^4 \frac{x}{2} \]
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\[ y’ = \mathrm{e}^{ax} \left[ a \cosh bx + b \text{ senh } bx \right] \]
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\[ y’ = \frac{1}{2} \sqrt[4]{ \frac{1 + \tanh x}{ 1 – \tanh x } } = \frac{1}{ 2 \sqrt{ \cosh x – \text{ senh } x } } \]\[ \begin{aligned} y’ &= \frac{1}{2} \sqrt[4]{ \frac{1 + \tanh x}{ 1 – \tanh x } } \\[.5em] &= \frac{1}{ 2 \sqrt{ \cosh x – \text{ senh } x } } \end{aligned} \]
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\[ y’ = \frac{ – 2 \text{ cosech}^{-1} x }{ \mid x \mid \sqrt{ 1 + x^2 } } \]
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\[ y’ = \frac{2x}{\sqrt{x^4 + a^4}} \]
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\[ y’ = -\text{cosec} x \]
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\[ y’ = \frac{2}{ 1 – x^4 } \]