Answer is: option3

Solution:

Problem: We are given the equation: \[ 2x^4 - xy + 3y^3 = 12 \] and we need to find \( \frac{dy}{dx} \) in terms of \( x \) and \( y \).

Step 1: Differentiate the equation implicitly

The equation is:

\[ 2x^4 - xy + 3y^3 = 12 \] Differentiate both sides with respect to \( x \) (implicit differentiation):

• For \( 2x^4 \), the derivative is: \[ \frac{d}{dx}(2x^4) = 8x^3 \]
• For \( -xy \), apply the product rule: \[ \frac{d}{dx}(-xy) = -\left( \frac{d}{dx}(x)y + x\frac{dy}{dx} \right) = -\left( y + x\frac{dy}{dx} \right) \]
• For \( 3y^3 \), apply the chain rule: \[ \frac{d}{dx}(3y^3) = 9y^2 \frac{dy}{dx} \]
• For the constant \( 12 \), the derivative is: \[ \frac{d}{dx}(12) = 0 \]

Therefore, the differentiated equation is: \[ 8x^3 - \left( y + x\frac{dy}{dx} \right) + 9y^2 \frac{dy}{dx} = 0 \]

Step 2: Solve for \( \frac{dy}{dx} \)

Now, let's solve for \( \frac{dy}{dx} \). First, expand the equation:

\[ 8x^3 - y - x\frac{dy}{dx} + 9y^2 \frac{dy}{dx} = 0 \]

Group the terms involving \( \frac{dy}{dx} \):

\[ - x\frac{dy}{dx} + 9y^2 \frac{dy}{dx} = y - 8x^3 \]

Factor out \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} (-x + 9y^2) = y - 8x^3 \]

Finally, solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{y - 8x^3}{9y^2 - x} \]

Final Answer:

The expression for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{y - 8x^3}{9y^2 - x} \]

Thus, the correct answer is Option C.