To solve the problem using implicit differentiation, we need to find $\frac{dy}{dx}$ for the equation:

$e^{x^3y} = 6x + 5y + 3$

Step 1: Differentiate both sides with respect to $x$.

We'll use the chain rule and the product rule where necessary. Let’s start:

1. Left-hand side: $\frac{d}{dx} \left( e^{x^3y} \right) = e^{x^3y} \cdot \frac{d}{dx}(x^3y)$ Using the product rule for $x^3y$: $\frac{d}{dx}(x^3y) = 3x^2y + x^3 \frac{dy}{dx}$ So, the derivative of the left-hand side becomes: $e^{x^3y} \cdot (3x^2y + x^3 \frac{dy}{dx})$

2. Right-hand side: Differentiate $6x + 5y + 3$ term-by-term: $\frac{d}{dx}(6x + 5y + 3) = 6 + 5 \frac{dy}{dx}$

Step 2: Write the full equation.

Combining both sides: $e^{x^3y} \cdot (3x^2y + x^3 \frac{dy}{dx}) = 6 + 5 \frac{dy}{dx}$

Step 3: Solve for $\frac{dy}{dx}$.

Expand the left-hand side: $e^{x^3y} \cdot 3x^2y + e^{x^3y} \cdot x^3 \frac{dy}{dx} = 6 + 5 \frac{dy}{dx}$

Rearrange terms to isolate $\frac{dy}{dx}$: $e^{x^3y} \cdot x^3 \frac{dy}{dx} - 5 \frac{dy}{dx} = 6 - e^{x^3y} \cdot 3x^2y$

Factor out $\frac{dy}{dx}$ on the left: $\left(e^{x^3y} \cdot x^3 - 5\right) \frac{dy}{dx} = 6 - e^{x^3y} \cdot 3x^2y$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{6 - e^{x^3y} \cdot 3x^2y}{e^{x^3y} \cdot x^3 - 5}$

Final Answer:

$\frac{dy}{dx} = \frac{6 - 3x^2y e^{x^3y}}{x^3 e^{x^3y} - 5}$

Let me know if you'd like further clarification or additional details!

Related Questions:

1. How does implicit differentiation work with exponential functions?
2. What is the significance of the chain rule in implicit differentiation?
3. How would the solution change if $y$ were not multiplied by $x^3$?
4. Can we verify this result by substituting values for $x$ and $y$?
5. How do logarithmic properties simplify exponential derivatives?

Tip:

Always double-check your derivative when dealing with composite functions like $e^{x^3y}$; the chain rule is critical.