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Find the derivative use logarithmic differentiation \( y=\sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \)

Ask by Grant Schneider. in the United States
Mar 17,2025

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The derivative of \( y = \sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \) is \[ \frac{dy}{dx} = \frac{2}{3} \left(\frac{x^{2}+4}{x+1}\right)^{\frac{2}{3}} \left(\frac{2x}{x^{2}+4} - \frac{1}{x+1}\right). \]

Solución

We start with \[ y=\left(\frac{x^{2}+4}{x+1}\right)^{\frac{2}{3}}. \] Taking the natural logarithm of both sides gives \[ \ln y = \frac{2}{3} \ln\left(\frac{x^{2}+4}{x+1}\right). \] Using the logarithm property \(\ln \frac{a}{b}=\ln a - \ln b\), we have \[ \ln y = \frac{2}{3}\left[\ln\left(x^{2}+4\right) - \ln\left(x+1\right)\right]. \] Differentiate both sides with respect to \(x\). For the left-hand side, use the chain rule: \[ \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx} \ln y. \] For the right-hand side, differentiate term by term: \[ \frac{d}{dx}\ln\left(x^{2}+4\right) = \frac{2x}{x^{2}+4}, \quad \frac{d}{dx}\ln\left(x+1\right) = \frac{1}{x+1}. \] Thus, \[ \frac{1}{y}\frac{dy}{dx}=\frac{2}{3}\left(\frac{2x}{x^{2}+4} - \frac{1}{x+1}\right). \] Multiplying both sides by \(y\) yields \[ \frac{dy}{dx}= y \cdot \frac{2}{3}\left(\frac{2x}{x^{2}+4} - \frac{1}{x+1}\right). \] Substituting back \(y=\left(\frac{x^{2}+4}{x+1}\right)^{\frac{2}{3}}\), we obtain \[ \frac{dy}{dx} = \frac{2}{3}\left(\frac{x^{2}+4}{x+1}\right)^{\frac{2}{3}} \left(\frac{2x}{x^{2}+4} - \frac{1}{x+1}\right). \]

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To differentiate the function \( y=\sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \) using logarithmic differentiation, we first take the natural logarithm of both sides: \[ \ln(y) = \ln\left(\sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}}\right) \] Using properties of logarithms, we can simplify the right side: \[ \ln(y) = \frac{1}{3} \ln\left(\left(\frac{x^{2}+4}{x+1}\right)^{2}\right) = \frac{2}{3} \ln\left(\frac{x^{2}+4}{x+1}\right) \] Now we apply the quotient rule for logarithms: \[ \ln(y) = \frac{2}{3} \left( \ln(x^2 + 4) - \ln(x + 1) \right) \] Next, we differentiate both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{3} \left( \frac{2x}{x^2 + 4} - \frac{1}{x + 1} \right) \] Now multiplying both sides by \( y \) and substituting back for \( y \): \[ \frac{dy}{dx} = y \cdot \frac{2}{3} \left( \frac{2x}{x^2 + 4} - \frac{1}{x + 1} \right) \] Substituting \( y = \sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \): \[ \frac{dy}{dx} = \sqrt[3]{\left(\frac{x^{2}+4}{x+1}\right)^{2}} \cdot \frac{2}{3} \left( \frac{2x}{x^2 + 4} - \frac{1}{x + 1} \right) \] This is the derivative of \( y \) using logarithmic differentiation.

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If you invest \( \$ 1000 \) for \( t \) years at \( 6.71 \% \) compounded at the frequencies below, find the following. (a) Suppose you compound at \( 6.71 \% \) monthly. i) Report an expression equivalent to the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded monthly by completing the box with the growth factor if compounded annually. 1000 \( \square \) Number \( t \) (Round to \( \underline{4} \) decimal places.) ii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (b) Suppose you compound at \( 6.71 \% \) continuously. i) You would expect \( 6.71 \% \) compounded continuously to give a \( \square \) Click for List yield than what is given in part (a). ii) Complete the boxes below to report the expression for the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded continuously and the equivalent growth factor if compounded annually. \[ \begin{array}{l} 1000 e^{(\text {Number } t)} \\ \approx 1000(\text { Number })^{t} \end{array} \] (Round to \( \underline{4} \) decimal places.) iii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (c) Complete the boxes to summarize: i) From part (a) we have that 6.71 \% compounded monthly is equivalent to \( \square \) Number \( \% \) compounded annually. ii) From part (b) we have that 6.71 \% compounded continuously is equivalent to \( \square \) Number \( \% \) compounded annually.
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