find dy/dx by implicit differentiation. e^x/y = 4x - y y =

Explanation:

Step1: Differentiate both sides

Differentiate $e^{x/y}$ and $4x - y$ with respect to $x$. For $e^{x/y}$, use the chain - rule and quotient - rule. The derivative of $e^{u}$ with respect to $x$ is $e^{u}\cdot u'$, where $u=\frac{x}{y}$. The derivative of $4x - y$ with respect to $x$ is $4 - y'$. The derivative of $u = \frac{x}{y}$ using the quotient - rule $\left(\frac{v}{w}\right)'=\frac{v'w - vw'}{w^{2}}$ (where $v = x$, $v'=1$, $w = y$, $w'=y'$) is $\frac{1\cdot y-x\cdot y'}{y^{2}}$. So the derivative of $e^{x/y}$ is $e^{x/y}\cdot\frac{y - xy'}{y^{2}}$. $$e^{x/y}\cdot\frac{y - xy'}{y^{2}}=4 - y'$$

Step2: Expand and collect terms with $y'$

Expand the left - hand side: $\frac{e^{x/y}y}{y^{2}}-\frac{e^{x/y}xy'}{y^{2}}=4 - y'$. $\frac{e^{x/y}}{y}-\frac{e^{x/y}x}{y^{2}}y'=4 - y'$. Move all terms with $y'$ to one side: $y'-\frac{e^{x/y}x}{y^{2}}y'=4-\frac{e^{x/y}}{y}$. Factor out $y'$: $y'\left(1 - \frac{e^{x/y}x}{y^{2}}\right)=4-\frac{e^{x/y}}{y}$.

Step3: Solve for $y'$

$y'=\frac{4-\frac{e^{x/y}}{y}}{1 - \frac{e^{x/y}x}{y^{2}}}$. Multiply the numerator and denominator by $y^{2}$ to simplify: $y'=\frac{4y^{2}-ye^{x/y}}{y^{2}-xe^{x/y}}$

Answer:

$\frac{4y^{2}-ye^{x/y}}{y^{2}-xe^{x/y}}$