To find $\frac{dy}{dx}$ for the equation given:

$x^3 \cdot y + 5 \cdot x \cdot y^5 = 3$

we'll use implicit differentiation with respect to $x$.

Step-by-Step Solution

1. Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^3 \cdot y + 5 \cdot x \cdot y^5) = \frac{d}{dx}(3)$

2. Apply the product rule on each term of the left side. For the first term $x^3 \cdot y$: $\frac{d}{dx}(x^3 \cdot y) = \frac{d}{dx}(x^3) \cdot y + x^3 \cdot \frac{dy}{dx} = 3x^2 \cdot y + x^3 \cdot \frac{dy}{dx}$

3. Apply the product rule on the second term $5 \cdot x \cdot y^5$: $\frac{d}{dx}(5 \cdot x \cdot y^5) = 5 \cdot \frac{d}{dx}(x) \cdot y^5 + 5 \cdot x \cdot \frac{d}{dx}(y^5)$

   • Differentiate $x$ to get $1$, so the first part is $5 \cdot y^5$.
   • Differentiate $y^5$ to get $5y^4 \cdot \frac{dy}{dx}$, so the second part is $5 \cdot x \cdot 5y^4 \cdot \frac{dy}{dx} = 25xy^4 \cdot \frac{dy}{dx}$.

   Putting it together: $\frac{d}{dx}(5 \cdot x \cdot y^5) = 5y^5 + 25xy^4 \cdot \frac{dy}{dx}$

4. Substitute back into the equation: $3x^2 \cdot y + x^3 \cdot \frac{dy}{dx} + 5y^5 + 25xy^4 \cdot \frac{dy}{dx} = 0$

5. Isolate $\frac{dy}{dx}$: Combine all terms involving $\frac{dy}{dx}$: $x^3 \cdot \frac{dy}{dx} + 25xy^4 \cdot \frac{dy}{dx} = -3x^2 \cdot y - 5y^5$

6. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} (x^3 + 25xy^4) = -3x^2 \cdot y - 5y^5$

7. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-3x^2 \cdot y - 5y^5}{x^3 + 25xy^4}$

Final Answer

$\frac{dy}{dx} = \frac{-3x^2 \cdot y - 5y^5}{x^3 + 25xy^4}$

Would you like further details or have any questions on this?

Here are some related questions:

1. How does implicit differentiation work in finding derivatives for non-explicit functions?
2. What are some practical applications of implicit differentiation?
3. How does the product rule apply to functions of two variables like $x$ and $y$?
4. What would happen if we wanted to find the second derivative, $\frac{d^2y}{dx^2}$?
5. Why is it necessary to rearrange the equation to isolate $\frac{dy}{dx}$ in implicit differentiation?

Tip: When using implicit differentiation, remember to apply the product and chain rules carefully whenever $y$ is multiplied by functions of $x$.