Find dy/dx for the equation sin(x2 y3) + 3x3 y2 = 6 using implicit differentiation.

To find $\frac{dy}{dx}$ for the given equation $\sin(x^2 y^3) + 3x^3 y^2 = 6$, we will use implicit differentiation. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate each term with respect to $x$.

$$\frac{d}{dx}(\sin(x^2 y^3)) + \frac{d}{dx}(3x^3 y^2) = \frac{d}{dx}(6)$$

Step 2: Differentiate $\sin(x^2 y^3)$ using the chain rule.
Let $u = x^2 y^3$. Then $\frac{d}{dx}(\sin(u)) = \cos(u) \frac{du}{dx}$.
To find $\frac{du}{dx} = \frac{d}{dx}(x^2 y^3)$, we use the product rule: $\frac{d}{dx}(fg) = f'g + fg'$.
Here, $f = x^2$ and $g = y^3$.
$\frac{d}{dx}(x^2) = 2x$.
$\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$ (by chain rule).
So, $\frac{du}{dx} = (2x)(y^3) + (x^2)(3y^2 \frac{dy}{dx}) = 2xy^3 + 3x^2 y^2 \frac{dy}{dx}$.
Therefore, $\frac{d}{dx}(\sin(x^2 y^3)) = \cos(x^2 y^3) (2xy^3 + 3x^2 y^2 \frac{dy}{dx})$.

Step 3: Differentiate $3x^3 y^2$ using the product rule.
Here, $f = 3x^3$ and $g = y^2$.
$\frac{d}{dx}(3x^3) = 9x^2$.
$\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$ (by chain rule).
So, $\frac{d}{dx}(3x^3 y^2) = (9x^2)(y^2) + (3x^3)(2y \frac{dy}{dx}) = 9x^2 y^2 + 6x^3 y \frac{dy}{dx}$.

Step 4: Differentiate the constant term.
$\frac{d}{dx}(6) = 0$.

Step 5: Substitute these derivatives back into the equation from Step 1.

$$\cos(x^2 y^3) (2xy^3 + 3x^2 y^2 \frac{dy}{dx}) + 9x^2 y^2 + 6x^3 y \frac{dy}{dx} = 0$$

Step 6: Expand and rearrange the equation to solve for $\frac{dy}{dx}$.

$$2xy^3 \cos(x^2 y^3) + 3x^2 y^2 \cos(x^2 y^3) \frac{dy}{dx} + 9x^2 y^2 + 6x^3 y \frac{dy}{dx} = 0$$

Group terms containing $\frac{dy}{dx}$:

$$(3x^2 y^2 \cos(x^2 y^3) + 6x^3 y) \frac{dy}{dx} = -2xy^3 \cos(x^2 y^3) - 9x^2 y^2$$

Factor out $\frac{dy}{dx}$:

$$\frac{dy}{dx} (3x^2 y^2 \cos(x^2 y^3) + 6x^3 y) = -(2xy^3 \cos(x^2 y^3) + 9x^2 y^2)$$

Isolate $\frac{dy}{dx}$:

$$$$