If `(cos^4 alpha)/(x ) + ( sin^4 alpha)/(y) =(1)/(x+y)` then ` (dy)/(dx)=`

Options

A. `xy`
B. `tan^2 alpha`
C. `0`
D. `(x^2 +y^2) sec^2 alpha`

Answer

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

\[ \frac{\cos^4 \alpha}{x} + \frac{\sin^4 \alpha}{y} = \frac{1}{x+y} \]

### Step 1: Multiply both sides by \(x + y\)

To eliminate the fraction, we multiply both sides of the equation by \(x + y\):

\[ (x + y) \left( \frac{\cos^4 \alpha}{x} + \frac{\sin^4 \alpha}{y} \right) = 1 \]

### Step 2: Distribute \(x + y\)

Distributing \(x + y\) across the terms gives:

\[ y \cos^4 \alpha + x \sin^4 \alpha = 1 \]

### Step 3: Rearranging the equation

We can rearrange the equation to isolate the terms involving \(x\) and \(y\):

\[ y \cos^4 \alpha + x \sin^4 \alpha - 1 = 0 \]

### Step 4: Differentiate implicitly

Now, we differentiate both sides of the equation with respect to \(x\):

\[ \frac{d}{dx}(y \cos^4 \alpha) + \frac{d}{dx}(x \sin^4 \alpha) - \frac{d}{dx}(1) = 0 \]

Using the product rule on the first term and the second term:

\[ \cos^4 \alpha \frac{dy}{dx} + y \cdot 0 + \sin^4 \alpha + x \cdot 0 = 0 \]

This simplifies to:

\[ \cos^4 \alpha \frac{dy}{dx} + \sin^4 \alpha = 0 \]

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

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

\[ \cos^4 \alpha \frac{dy}{dx} = -\sin^4 \alpha \]

\[ \frac{dy}{dx} = -\frac{\sin^4 \alpha}{\cos^4 \alpha} \]

### Step 6: Simplify the expression

Using the identity \(\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha}\):

\[ \frac{dy}{dx} = -\tan^4 \alpha \]

### Final Answer

Thus, the final answer is:

\[ \frac{dy}{dx} = -\tan^4 \alpha \]

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