18. [0/1 Points] DETAILS MY NOTES SCALCET9 3.5.021. PREVIOUS ANSWERS ASK YOUR TEACHER Find $$ \frac{d y}{d x} $$ by implicit differentiation. $$ \begin{array}{c} e^{x / y}=5 x-y \ \frac{d y}{d x}=\frac{5 y^{2}-e^{x}}{y^{2}-\frac{e^{x}}{y^{x}}} \end{array} $$ Need Help? Read It Watch It

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We start with the equation

$$
e^{x/y} = 5x - y.
$$

Differentiate both sides with respect to $$ x $$.

1. Differentiate the left side using the chain rule. Write
   $$
   \frac{d}{dx}\left(e^{x/y}\right) = e^{x/y} \cdot \frac{d}{dx}\left(\frac{x}{y}\right).
   $$
   To differentiate $$ \frac{x}{y} $$, use the quotient rule:
   $$
   \frac{d}{dx}\left(\frac{x}{y}\right) = \frac{y\cdot 1 - x\cdot\frac{dy}{dx}}{y^2} = \frac{y - x\,y'}{y^2},
   $$
   where we write $$ y'=\frac{dy}{dx} $$. So the derivative of the left side becomes
   $$
   e^{x/y}\cdot\frac{y-x\,y'}{y^2}.
   $$

2. Differentiate the right side. Since
   $$
   \frac{d}{dx}(5x-y)=5-y',
   $$
   we have the differentiated equation:
   $$
   e^{x/y}\frac{y-x\,y'}{y^2}=5-y'.
   $$

3. Multiply both sides by $$ y^2 $$ to clear the denominator:
   $$
   e^{x/y}(y-x\,y')=y^2(5-y').
   $$

4. Distribute on both sides:
   $$
   e^{x/y}y - e^{x/y}x\,y' = 5y^2 - y^2y'.
   $$

5. Group the terms involving $$ y' $$ on one side:
   $$
   -e^{x/y}x\,y' + y^2y' = 5y^2 - e^{x/y}y.
   $$

6. Factor out $$ y' $$ on the left:
   $$
   y'\left(y^2-e^{x/y}x\right)=5y^2-e^{x/y}y.
   $$

7. Finally, solve for $$ y' $$ by dividing:
   $$
   y' = \frac{5y^2 - e^{x/y}y}{y^2 - e^{x/y}x}.
   $$

Thus, the derivative is

$$
\frac{dy}{dx} = \frac{5y^2 - y\,e^{x/y}}{y^2 - x\,e^{x/y}}.
$$
$$ \frac{dy}{dx} = \frac{5y^2 - y\,e^{x/y}}{y^2 - x\,e^{x/y}} $$

[0/1 Points]
DETAILS
MY NOTES
SCALCET9 3.5.021.
PREVIOUS ANSWERS
ASK YOUR TEACHER

Find by implicit differentiation.

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[0/1 Points] DETAILS MY NOTES SCALCET9 3.5.021. PREVIOUS ANSWERS ASK YOUR TEACHER Find by implicit differentiation. Need Help? Read It Watch It