Question

use implicit differentiation to find $\frac{dy}{dx}$. $ysinleft(\frac{1}{y}
ight)=1 - 7xy$. $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $y\sin(\frac{1}{y})$ and $1 - 7xy$ with respect to $x$.
For the left - hand side, use the product rule $(uv)^\prime=u^\prime v + uv^\prime$ where $u = y$ and $v=\sin(\frac{1}{y})$.
The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$, and for $\sin(\frac{1}{y})$, use the chain - rule. Let $u=\frac{1}{y}$, then $\frac{d}{dx}\sin(\frac{1}{y})=\cos(\frac{1}{y})\cdot(-\frac{1}{y^{2}})\frac{dy}{dx}$.
So, $\frac{d}{dx}(y\sin(\frac{1}{y}))=\frac{dy}{dx}\sin(\frac{1}{y})+y\cos(\frac{1}{y})\cdot(-\frac{1}{y^{2}})\frac{dy}{dx}=\frac{dy}{dx}\sin(\frac{1}{y})-\frac{1}{y}\cos(\frac{1}{y})\frac{dy}{dx}$.
For the right - hand side, the derivative of the constant $1$ is $0$, and for $-7xy$, use the product rule: $\frac{d}{dx}(-7xy)=-7y-7x\frac{dy}{dx}$.

Step2: Solve for $\frac{dy}{dx}$

Set the derivatives of both sides equal:
$\frac{dy}{dx}\sin(\frac{1}{y})-\frac{1}{y}\cos(\frac{1}{y})\frac{dy}{dx}=-7y - 7x\frac{dy}{dx}$.
Factor out $\frac{dy}{dx}$ on the left - hand side:
$\frac{dy}{dx}(\sin(\frac{1}{y})-\frac{1}{y}\cos(\frac{1}{y}))=-7y - 7x\frac{dy}{dx}$.
Move all terms with $\frac{dy}{dx}$ to one side:
$\frac{dy}{dx}(\sin(\frac{1}{y})-\frac{1}{y}\cos(\frac{1}{y}) + 7x)=-7y$.
Then, $\frac{dy}{dx}=\frac{-7y}{\sin(\frac{1}{y})-\frac{1}{y}\cos(\frac{1}{y})+7x}$.

Answer:

$\frac{-7y}{\sin(\frac{1}{y})-\frac{1}{y}\cos(\frac{1}{y})+7x}$