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

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}
$