`tan^(-1)(x^(2)+y^(2)) = a`

Explanation

To solve the problem \( \tan^{-1}(x^2 + y^2) = a \), we will differentiate both sides with respect to \( x \).

### Step-by-Step Solution:

1. **Differentiate Both Sides:**
We start with the equation:
\[ \tan^{-1}(x^2 + y^2) = a \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(\tan^{-1}(x^2 + y^2)) = \frac{d}{dx}(a) \]

2. **Apply the Chain Rule:**
The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \). Here, \( u = x^2 + y^2 \):
\[ \frac{d}{dx}(\tan^{-1}(x^2 + y^2)) = \frac{1}{1 + (x^2 + y^2)^2} \cdot \frac{d}{dx}(x^2 + y^2) \]
Since \( a \) is a constant, its derivative is 0:
\[ \frac{d}{dx}(a) = 0 \]

3. **Differentiate \( x^2 + y^2 \):**
Now we differentiate \( x^2 + y^2 \):
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 2x + 2y \frac{dy}{dx} \]

4. **Combine the Results:**
Substituting back into our differentiation:
\[ \frac{1}{1 + (x^2 + y^2)^2} \cdot (2x + 2y \frac{dy}{dx}) = 0 \]

5. **Setting the Equation to Zero:**
For the product to equal zero, we can set the term inside the parentheses to zero (since the fraction cannot be zero):
\[ 2x + 2y \frac{dy}{dx} = 0 \]

6. **Solve for \( \frac{dy}{dx} \):**
Rearranging gives:
\[ 2y \frac{dy}{dx} = -2x \]
Dividing both sides by \( 2y \):
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Final Result:
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{x}{y} \]