Derivación logarítmica
Utilizando la técnica de la derivación logarítmica hallar la derivada de las siguientes funciones:
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\[ y = x^{x^3} \]
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\[ y = x^{ \sqrt{x} }, \; x>0 \]
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\[ y = x^{\ln x}, \; x > 0 \]
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\[ y = ( \ln x )^{\ln x} \]
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\[ y = 2^{3^x} \]
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\[ y = a^x x^a \]
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\[ y = \sqrt[x]{ x } \]
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\[ y = \left( x^2 + 1 \right)^{\text{ sen } x} \]
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\[ y = \left( \text{ sen } x \right)^{ \cos x } \]
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\[ y = \left( 1 + \frac{1}{x} \right)^x \]
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\[ y = \frac{ x \left( x^2 - 1 \right) }{ \sqrt{ x^2 + 1 } } \]
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\[ y = \sqrt[3]{ \frac{x \left( x^2 - 1 \right) }{ (x + 1)^2 } } \]
Ver Respuestas
Respuestas
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\[ y’ = x^{x^3 + 2} ( 1 + 3 \ln x ) \]
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\[ y’ = \frac{1}{2} x^{ \sqrt{x} – \frac{1}{2} } (2 + \ln x) \]
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\[ y’ = 2 \ln x \cdot x^{ \ln x – 1 } \]
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\[ y’ = \frac{1}{x} (\ln x)^{\ln x} ( 1 + \ln (\ln x) ) \]
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\[ y’ = (\ln 2) ( \ln 3 ) 3^x 2^{ 3^x } \]
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\[ y’ = a^x x^a \left( \frac{a}{x} + \ln a \right) \]
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\[ y’ = \sqrt[x]{ x } \left( \frac{1 – \ln x}{x^2} \right) \]
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\[ y’ = \left( x^2 + 1 \right)^{\text{ sen } x} \left( \frac{2x \text{ sen } x}{ x^2 + 1} + \cos x \ln \left( x^2 + 1 \right) \right) \]\[ \begin{aligned} y’ =& \left( x^2 + 1 \right)^{\text{ sen } x} \\[.5em] & \times \left( \frac{2x \text{ sen } x}{ x^2 + 1} + \cos x \ln \left( x^2 + 1 \right) \right) \end{aligned} \]
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\[ y’ = (\text{ sen } x)^{\cos x} \left( \frac{ \cos^2 x }{ \text{ sen } x } – \text{ sen } x \ln (\text{ sen } x) \right) \]\[ \begin{aligned} y’ =& (\text{ sen } x)^{\cos x} \\[.5em] &\hspace{1em} \times \left( \frac{ \cos^2 x }{ \text{ sen } x } – \text{ sen } x \ln (\text{ sen } x) \right) \end{aligned} \]
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\[ y’ = \left( 1 + \frac{1}{x} \right)^x \left( \ln \frac{x + 1}{x} – \frac{1}{ x + 1 } \right) \]\[ \begin{aligned} y’ =& \left( 1 + \frac{1}{x} \right)^x \\[.5em] &\hspace{2em} \times \left( \ln \frac{x + 1}{x} – \frac{1}{ x + 1 } \right) \end{aligned} \]
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\[ y’ = \frac{x \left( x^2 – 1 \right)}{ \sqrt{ x^2 + 1 } } \left( \frac{1}{x} + \frac{2x}{ x^2 – 1 } – \frac{x}{x^2 + 1} \right) \]\[ \begin{aligned} y’ =& \frac{x \left( x^2 – 1 \right)}{ \sqrt{ x^2 + 1 } } \\[.5em] &\hspace{.5em} \times \left( \frac{1}{x} + \frac{2x}{ x^2 – 1 } – \frac{x}{x^2 + 1} \right) \end{aligned} \]
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\[ y’ = \frac{1}{3} \sqrt[3]{ \frac{ x \left( x^2 – 1 \right) }{ (x + 1)^2 }} \left( \frac{1}{x} + \frac{2x}{ x^2 – 1 } – \frac{2}{x + 1} \right) \]\[ \begin{aligned} y’ =& \frac{1}{3} \sqrt[3]{ \frac{ x \left( x^2 – 1 \right) }{ (x + 1)^2 }} \\[.5em] &\hspace{1em} \times \left( \frac{1}{x} + \frac{2x}{ x^2 – 1 } – \frac{2}{x + 1} \right) \end{aligned} \]