Question

use implicit differentiation to find $\frac{dy}{dx}$. $2ysin(\frac{1}{y}) = 8 - 9xy$ $\frac{dy}{dx}=square$

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 $v=\sin(\frac{1}{y})$, using chain - rule, let $t=\frac{1}{y}=y^{-1}$, then $\frac{dv}{dt}=\cos(t)$ and $\frac{dt}{dy}=-y^{-2}=-\frac{1}{y^{2}}$. So $\frac{dv}{dy}=\cos(\frac{1}{y})(-\frac{1}{y^{2}})$.
The derivative of the left - hand side is $2\frac{dy}{dx}\sin(\frac{1}{y})+2y\cos(\frac{1}{y})(-\frac{1}{y^{2}})\frac{dy}{dx}=2\frac{dy}{dx}\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})\frac{dy}{dx}$.

Step2: Differentiate right - hand side

Differentiate $8 - 9xy$ with respect to $x$. The derivative of a constant $8$ is $0$, and for $-9xy$, using product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=-9x$ and $v = y$.
$u^\prime=-9$ and $v^\prime=\frac{dy}{dx}$, so the derivative of $-9xy$ is $-9y-9x\frac{dy}{dx}$.

Step3: Set derivatives equal

Set the derivative of the left - hand side equal to the derivative of the right - hand side:
$2\frac{dy}{dx}\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})\frac{dy}{dx}=-9y - 9x\frac{dy}{dx}$.

Step4: Solve for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})+9x)=-9y$.
Then $\frac{dy}{dx}=\frac{-9y}{2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})+9x}$.

Answer:

$\frac{-9y}{2\sin(\frac{1}{y})-\frac{2}{y}\cos(\frac{1}{y})+9x}$