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Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 42

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Find the derivative of the function $y = \sqrt{x} e^{x^2} (x^2 + 1)^{10}$, using log differentiation

$ \begin{equation} \begin{aligned} \ln y &= \ln \left[ \sqrt{x} e^{x^2} (x^2 + 1)^{10} \right]\\ \\ \ln y &= \ln \sqrt{x} + \ln e^{x^2} + \ln (x^2 + 1)^{10}\\ \\ \ln y &= \ln x^{\frac{1}{2}} + x^2 + \ln (x^2 +1)^{10}\\ \\ \ln y &= \frac{1}{2} \ln x + x^2 + 10 \ln (x^2 +1)\\ \\ \frac{d}{dx} \ln y &= \frac{1}{2} \frac{d}{dx} (\ln x) + \frac{d}{dx}(x^2) + 10 \frac{d}{dx} \ln (x^2 +1)\\ \\ \frac{1}{y}\frac{dy}{dx} &= \frac{1}{2} \cdot \frac{1}{x} + 2x + 10 \cdot \frac{1}{x^2 + 1} \frac{d}{dx} (x^2 +1)\\ \\ \frac{1}{y} y' &= \frac{1}{2x} + 2x + \frac{10}{x^2 +1} \cdot 2x\\ \\ \frac{y'}{y} &= \frac{1}{2x} + 2x + \frac{20x}{x^2 + 1}\\ \\ y' &= y \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right)\\ \\ y' &= \left( \sqrt{x} e^{x^2} (x^2 + 1)^{10} \right) \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \end{aligned} \end{equation} $