Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 42

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Determine the derivative of $\displaystyle y = x^2 \sin h^{-1} (2x)$. Simplify where possible.

$ \begin{equation} \begin{aligned} y' =& \frac{d}{dx} [x^2 \sin h^{-1} (2x)] \\ \\ y' =& x^2 \cdot \frac{d}{dx} [\sin h^{-1} (2x)] + \sin h^{-1} (2x) \cdot \frac{d}{dx} (x^2) \\ \\ y' =& x^2 \cdot \frac{1}{\sqrt{(2x)^2 + 1}} \cdot \frac{d}{dx} (2x) + \sin h^{-1} (2x) \cdot 2x \\ \\ y' =& \frac{x^2}{\sqrt{4x^2 + 1}} \cdot 2 + 2x \sin h^{-1} (2x) \\ \\ y' =& \frac{2x^2}{\sqrt{4x^2 + 1}} + 2 x \sin h^{-1} (2x) \end{aligned} \end{equation} $