Question

use implicit differentiation to find $\frac{dy}{dx}$
$2ysinleft(\frac{1}{y}
ight)=2 - xy$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate left - hand side

Differentiate $2y\sin(\frac{1}{y})$ using product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 2y$ and $v=\sin(\frac{1}{y})$.
$u^\prime=2\frac{dy}{dx}$, and for $v=\sin(\frac{1}{y})$, by chain - rule, $v^\prime=\cos(\frac{1}{y})\times(-\frac{1}{y^{2}})\frac{dy}{dx}$.
So, $\frac{d}{dx}(2y\sin(\frac{1}{y}))=2\frac{dy}{dx}\sin(\frac{1}{y})+2y\cos(\frac{1}{y})\times(-\frac{1}{y^{2}})\frac{dy}{dx}=2\frac{dy}{dx}\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})\frac{dy}{dx}$.

Step2: Differentiate right - hand side

Differentiate $2-xy$ with respect to $x$. $\frac{d}{dx}(2) = 0$, and using product rule for $xy$ where $u = x$ and $v = y$, $(xy)^\prime=y + x\frac{dy}{dx}$. So, $\frac{d}{dx}(2-xy)=-y - x\frac{dy}{dx}$.

Step3: Set derivatives equal

Set $\frac{d}{dx}(2y\sin(\frac{1}{y}))=\frac{d}{dx}(2 - xy)$.
$2\frac{dy}{dx}\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})\frac{dy}{dx}=-y - x\frac{dy}{dx}$.

Step4: Isolate $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y}))=-y - x\frac{dy}{dx}$.
Move terms with $\frac{dy}{dx}$ to one side: $\frac{dy}{dx}(2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})+x)=-y$.
Then $\frac{dy}{dx}=\frac{-y}{2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})+x}$.

Answer:

$\frac{-y}{2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})+x}$