Single Variable Calculus

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Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 48

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Find $y'$ if $\ln x y = y \sin x$.

$ \begin{equation} \begin{aligned} & \text{if } y = \ln xy = y \sin x, \text{ then by taking the derivative implicitly we have.. } \\ \\ & \frac{\displaystyle \frac{d}{dx} (xy) }{xy} = \frac{d}{dx} (y \sin x) \\ \\ & \frac{\displaystyle x \cdot \frac{d}{dy} (y) \frac{dy}{dx} + y(1) }{xy} = y \cos x + \sin x \cdot \frac{d}{dy} (y) \frac{dy}{dx} \\ \\ & \frac{\displaystyle x \frac{dy}{dx} + y}{xy} = y \cos x + \sin x \frac{dy}{dx} \\ \\ & x \frac{dy}{dx} + y = xy^2 \cos x+ xy \sin x \frac{dy}{dx} \\ \\ & y - xy^2 \cos x = xy \sin x \frac{dy}{dx} - x \frac{dy}{dx} \\ \\ & y (1 - xy \cos x) = x (y \sin x - 1) \frac{dy}{dx} \\ \\ & \frac{dy}{dx} = \frac{y (1 - xy \cos x)}{x (y \sin x - 1)} \end{aligned} \end{equation} $

Recall that the first derivative is equal to the slope of the tangent line at some point.

Thus, at point $(2, 0)$,

$ \begin{equation} \begin{aligned} y' =& \frac{3(2)^2}{2^3 - 7} \\ \\ y' =& 12 \end{aligned} \end{equation} $

Therefore, the equation of the tangent line to the curve can be determined by using the point slope form.

$ \begin{equation} \begin{aligned} y - y_1 =& m (x - x_1) \\ \\ y - 0 =& 12 (x - 2) \\ \\ y =& 12 x - 24 \end{aligned} \end{equation} $

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