If ` x ^(2)y^(2) =tan ^(-1) sqrt(x^(2) +y^(2) )+cot ^(-1) sqrt(x^(2) +y^(2)),then (dy)/(dx)=`

Options

A. ` (-x)/( y)`
B. ` (x)/(y)`
C. ` (-y)/(x)`
D. ` (y)/(x)`

Answer

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

\[ x^2 y^2 = \tan^{-1}(\sqrt{x^2 + y^2}) + \cot^{-1}(\sqrt{x^2 + y^2}) \]

### Step 1: Use the Identity
We know from trigonometric identities that:

\[ \tan^{-1}(p) + \cot^{-1}(p) = \frac{\pi}{2} \]

for any \( p \). In our case, let \( p = \sqrt{x^2 + y^2} \). Therefore, we can rewrite the equation as:

\[ x^2 y^2 = \frac{\pi}{2} \]

### Step 2: Rearranging the Equation
From the equation \( x^2 y^2 = \frac{\pi}{2} \), we can express it in terms of \( xy \):

\[ xy = \sqrt{\frac{\pi}{2}} \]

### Step 3: Differentiate Both Sides
Now, we differentiate both sides with respect to \( x \). Using the product rule on the left side:

\[ \frac{d}{dx}(xy) = y \frac{dx}{dx} + x \frac{dy}{dx} = y + x \frac{dy}{dx} \]

Since the right side \( \sqrt{\frac{\pi}{2}} \) is a constant, its derivative is 0:

\[ \frac{d}{dx}\left(\sqrt{\frac{\pi}{2}}\right) = 0 \]

Thus, we have:

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

### Step 4: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:

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

Now, divide both sides by \( x \):

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

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:

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

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