To find $\frac{dy}{dx}$ using implicit differentiation from the equation:

$x^5 + y^5 = 15xy$

Step-by-Step Solution:

1. Differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$. Remember to apply the product rule and chain rule where necessary.

   $\frac{d}{dx}(x^5) + \frac{d}{dx}(y^5) = \frac{d}{dx}(15xy)$

2. Differentiate each term:

   • The derivative of $x^5$ is: $5x^4$
   • The derivative of $y^5$ (using the chain rule since $y$ is a function of $x$) is: $5y^4 \frac{dy}{dx}$
   • For the right-hand side, use the product rule for $15xy$: $\frac{d}{dx}(15xy) = 15 \left( x \frac{dy}{dx} + y \right)$

3. Now, substitute these into the differentiated equation: $5x^4 + 5y^4 \frac{dy}{dx} = 15(x \frac{dy}{dx} + y)$

4. Rearrange the terms to collect all $\frac{dy}{dx}$ terms on one side: $5y^4 \frac{dy}{dx} - 15x \frac{dy}{dx} = 15y - 5x^4$

5. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(5y^4 - 15x) = 15y - 5x^4$

6. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{15y - 5x^4}{5y^4 - 15x}$

7. Simplify the expression if possible: $\frac{dy}{dx} = \frac{3y - x^4}{y^4 - 3x}$

This is the derivative $\frac{dy}{dx}$ using implicit differentiation.

Would you like further details or any clarifications?

Related Questions:

1. How do you apply the product rule in differentiation?
2. What is the chain rule, and why is it important in implicit differentiation?
3. Can you find the second derivative $\frac{d^2y}{dx^2}$ from this equation?
4. How would the process change if $y$ was a constant?
5. How does implicit differentiation differ from explicit differentiation?

Tip:

Always carefully apply the product and chain rules when differentiating terms involving both $x$ and $y$, as this helps to avoid common mistakes in implicit differentiation.