Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 44

Differentiate $\displaystyle y = \frac{(x^2 + 1)^4}{(2x + 1)^3 (3x - 1)^5}$


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\begin{equation}
\begin{aligned}

\ln y =& \ln \left[ \frac{(x^2 + 1)^4}{(2x + 1)^3 (3x - 1)^5} \right]
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\ln y =& \ln (x^2 + 1)^4 - [\ln (2x + 1)^3 + \ln (3x - 1)^5]
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\ln y =& 4 \ln (x^2 + 1) - 3 \ln (2x + 1) - 5 \ln (3x - 1)
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\frac{d}{dx} (\ln y) =& 4 \frac{d}{dx} [\ln (x^2 + 1)] - 3 \frac{d}{dx} [\ln (2x + 1)] - 5 \frac{d}{dx} [\ln (3x - 1)]
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\frac{1}{y} \frac{dy}{dx} =& 4 \cdot \frac{1}{x^2 + 1} \frac{d}{dx} (x^2 + 1) - 3 \cdot \frac{1}{2x + 1} \frac{d}{dx} (2x + 1) - 5 \cdot \frac{1}{3x - 1} \frac{d}{dx} (3x - 1)
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\frac{1}{y} y' =& \frac{4}{x^2 + 1} \cdot 2x - \frac{3}{2x + 1} \cdot 2 - \frac{5}{3x - 1} \cdot 3
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\frac{y'}{y} =& \frac{8x}{x^2 + 1} - \frac{6}{2x + 1} - \frac{15}{3x - 1}
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y' =& y \left( \frac{8x}{x^2 + 1} - \frac{6}{2x + 1} - \frac{15}{3x - 1} \right)
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y' =& \frac{(x^2 + 1)^4}{(2x + 1)^3 (3x - 1)^5} \left( \frac{8x}{x^2 + 1} - \frac{6}{2x + 1} - \frac{15}{3x - 1} \right)

\end{aligned}
\end{equation}
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