To find $y'$ (the derivative of $y$ with respect to $x$) by implicit differentiation, let's differentiate both sides of the equation with respect to $x$. The equation is:

$e^{x/y} = 3x - 4y$

Step 1: Differentiate the left-hand side

The derivative of $e^{x/y}$ requires the chain rule:

$\frac{d}{dx} \left( e^{x/y} \right) = e^{x/y} \cdot \frac{d}{dx} \left( \frac{x}{y} \right)$

Now, use the quotient rule to differentiate $\frac{x}{y}$:

$\frac{d}{dx} \left( \frac{x}{y} \right) = \frac{y \cdot 1 - x \cdot y'}{y^2} = \frac{y - x y'}{y^2}$

Thus, the derivative of the left-hand side is:

$e^{x/y} \cdot \frac{y - x y'}{y^2}$

Step 2: Differentiate the right-hand side

The derivative of $3x - 4y$ is:

$\frac{d}{dx}(3x - 4y) = 3 - 4y'$

Step 3: Combine the derivatives

Now, we combine the results of both sides:

$e^{x/y} \cdot \frac{y - x y'}{y^2} = 3 - 4y'$

Step 4: Solve for $y'$

First, multiply both sides by $y^2$ to simplify:

$e^{x/y} (y - x y') = y^2 (3 - 4y')$

Expand both sides:

$e^{x/y} y - e^{x/y} x y' = 3y^2 - 4y^2 y'$

Now, collect all the terms involving $y'$ on one side:

$- e^{x/y} x y' + 4y^2 y' = 3y^2 - e^{x/y} y$

Factor out $y'$:

$y'(- e^{x/y} x + 4y^2) = 3y^2 - e^{x/y} y$

Finally, solve for $y'$:

$y' = \frac{3y^2 - e^{x/y} y}{4y^2 - e^{x/y} x}$

This is the expression for $y'$ found by implicit differentiation.

Would you like further clarification or details?

Follow-up questions:

1. Can you explain the quotient rule used in this context?
2. What is the chain rule, and how was it applied to the term $e^{x/y}$?
3. Why did we factor out $y'$ in the final step?
4. How do you use implicit differentiation in other types of equations?
5. Can implicit differentiation be used for parametric equations?

Tip:

Always remember to use the chain rule when differentiating functions with variables inside an exponent!