Answer is: option4

\( \frac{3x^2 - 6xy + 3y^2}{ 3x^2 - 6xy + 3y^2 -\sec y \cdot \tan y } \)

Solution:

Problem: We are given the equation: \[ \sec(y) = (y - x)^3 \] and we need to find \( \frac{dy}{dx} \).

Step 1: Differentiate both sides of the equation with respect to \( x \)

The given equation is: \[ \sec(y) = (y - x)^3 \] To find \( \frac{dy}{dx} \), we differentiate both sides with respect to \( x \), applying implicit differentiation since \( y \) is a function of \( x \).

• For the left-hand side, the derivative of \( \sec(y) \) is: \[ \frac{d}{dx}(\sec(y)) = \sec(y) \tan(y) \frac{dy}{dx} \] (using the chain rule for the derivative of \( \sec(y) \)).
• For the right-hand side, we apply the chain rule to \( (y - x)^3 \): \[ \frac{d}{dx} \left( (y - x)^3 \right) = 3(y - x)^2 \cdot \frac{d}{dx}(y - x) \] Now, differentiate \( (y - x) \) with respect to \( x \): \[ \frac{d}{dx}(y - x) = \frac{dy}{dx} - 1 \] So the derivative of the right-hand side becomes: \[ 3(y - x)^2 \cdot (\frac{dy}{dx} - 1) \]

Thus, after differentiating both sides, we have the equation: \[ \sec(y) \tan(y) \frac{dy}{dx} = 3(y - x)^2 \cdot (\frac{dy}{dx} - 1) \]

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

Now, we solve this equation for \( \frac{dy}{dx} \). First, expand the right-hand side:

\[ \sec(y) \tan(y) \frac{dy}{dx} = 3(y - x)^2 \frac{dy}{dx} - 3(y - x)^2 \]

Move all terms involving \( \frac{dy}{dx} \) to one side:

\[ \sec(y) \tan(y) \frac{dy}{dx} - 3(y - x)^2 \frac{dy}{dx} = -3(y - x)^2 \]

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

\[ \frac{dy}{dx} \left( \sec(y) \tan(y) - 3(y - x)^2 \right) = -3(y - x)^2 \]

Now solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{-3(y - x)^2}{\sec(y) \tan(y) - 3(y - x)^2} \]

Final Answer:

The derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{-3(y - x)^2}{\sec(y) \tan(y) - 3(y - x)^2} \]