Pregunta
4. [-/1 Points] DETAILS MY NO Differentiate the function. \[ y=\frac{13}{\ln (x)} \] \( y^{\prime}=\square \)
Ask by Martin Klein. in the United States
Mar 20,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
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\( y' = -\frac{13}{x(\ln(x))^2} \)
Solución
Given the function
\[
y = \frac{13}{\ln(x)} = 13 \cdot (\ln(x))^{-1},
\]
we differentiate using the chain rule. The derivative of \((\ln(x))^{-1}\) is
\[
\frac{d}{dx} \left[(\ln(x))^{-1}\right] = -1 \cdot (\ln(x))^{-2} \cdot \frac{1}{x} = -\frac{1}{x(\ln(x))^2}.
\]
Thus, multiplying by 13 we obtain
\[
y' = 13 \cdot \left(-\frac{1}{x(\ln(x))^2}\right) = -\frac{13}{x(\ln(x))^2}.
\]
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Simplifique esta solución Mind Expander
To differentiate the function \( y = \frac{13}{\ln(x)} \), you can use the quotient rule or rewrite it as \( y = 13 (\ln(x))^{-1} \) and apply the chain rule. Using the quotient rule where \( u = 13 \) and \( v = \ln(x) \): \[ y' = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] Since \( \frac{du}{dx} = 0 \) and \( \frac{dv}{dx} = \frac{1}{x} \): \[ y' = \frac{\ln(x) \cdot 0 - 13 \cdot \frac{1}{x}}{(\ln(x))^2} = -\frac{13}{x (\ln(x))^2} \] So, the final answer is: \[ y' = -\frac{13}{x (\ln(x))^2} \]
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