Question

use implicit differentiation to find $\frac{dy}{dx}$
$3ysinleft(\frac{1}{y}
ight)=1 - 4xy$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate left - hand side

Differentiate $3y\sin(\frac{1}{y})$ using product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = 3y$ and $v=\sin(\frac{1}{y})$.
The derivative of $u = 3y$ with respect to $x$ is $3\frac{dy}{dx}$.
For $v=\sin(\frac{1}{y})$, use chain - rule. Let $t=\frac{1}{y}$, then $\frac{dv}{dt}=\cos(t)$ and $\frac{dt}{dx}=-\frac{1}{y^{2}}\frac{dy}{dx}$. So $\frac{dv}{dx}=\cos(\frac{1}{y})(-\frac{1}{y^{2}}\frac{dy}{dx})$.
The derivative of $3y\sin(\frac{1}{y})$ is $3\frac{dy}{dx}\sin(\frac{1}{y})+3y\cos(\frac{1}{y})(-\frac{1}{y^{2}}\frac{dy}{dx})=3\frac{dy}{dx}\sin(\frac{1}{y})-\frac{3}{y}\cos(\frac{1}{y})\frac{dy}{dx}$.

Step2: Differentiate right - hand side

Differentiate $1 - 4xy$ with respect to $x$. The derivative of a constant $1$ is $0$.
Using product rule for $-4xy$ where $u=-4x$ and $v = y$, we have $(-4x)^\prime y+(-4x)y^\prime=-4y-4x\frac{dy}{dx}$.

Step3: Set derivatives equal

Set the derivative of the left - hand side equal to the derivative of the right - hand side:
$3\frac{dy}{dx}\sin(\frac{1}{y})-\frac{3}{y}\cos(\frac{1}{y})\frac{dy}{dx}=-4y - 4x\frac{dy}{dx}$.

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

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(3\sin(\frac{1}{y})-\frac{3}{y}\cos(\frac{1}{y}) + 4x)=-4y$.
Then $\frac{dy}{dx}=\frac{-4y}{3\sin(\frac{1}{y})-\frac{3}{y}\cos(\frac{1}{y})+4x}$.

Answer:

$\frac{-4y}{3\sin(\frac{1}{y})-\frac{3}{y}\cos(\frac{1}{y}) + 4x}$