If ` sin ^(-1) (x^(3) +y^(3))= ` a then ` (dy)/(dx)= ` __________

Options

A. `(-x)/(cos a)`
B. `(-x^(2))/(y^(2))`
C. `(y^(2))//x^(2)`
D. `(sin a )/( y)`

Answer

To solve the problem, we start with the given equation:

\[ \sin^{-1}(x^3 + y^3) = a \]

We want to find \(\frac{dy}{dx}\).

### Step 1: Differentiate both sides with respect to \(x\)

Using implicit differentiation, we differentiate both sides:

\[ \frac{d}{dx}(\sin^{-1}(x^3 + y^3)) = \frac{d}{dx}(a) \]

Since \(a\) is a constant, its derivative is 0:

\[ \frac{d}{dx}(\sin^{-1}(x^3 + y^3)) = 0 \]

### Step 2: Apply the chain rule

Using the chain rule, we have:

\[ \frac{1}{\sqrt{1 - (x^3 + y^3)^2}} \cdot \frac{d}{dx}(x^3 + y^3) = 0 \]

### Step 3: Differentiate \(x^3 + y^3\)

Now, we differentiate \(x^3 + y^3\):

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

### Step 4: Substitute back into the equation

Substituting this back into our equation gives:

\[ \frac{1}{\sqrt{1 - (x^3 + y^3)^2}} \cdot (3x^2 + 3y^2 \frac{dy}{dx}) = 0 \]

Since the fraction is equal to zero, we can conclude that:

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

### Step 5: Solve for \(\frac{dy}{dx}\)

Rearranging the equation gives:

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

Dividing both sides by \(3y^2\) yields:

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

### Final Answer

Thus, we have:

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