Find `(dy)/(dx)` if, `tan^(-1)(x^2+y^2)=a`

Explanation

To find \(\frac{dy}{dx}\) given the equation \(\tan^{-1}(x^2 + y^2) = a\), we will differentiate both sides with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate Both Sides:**
Start with the equation:
\[ \tan^{-1}(x^2 + y^2) = a \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(\tan^{-1}(x^2 + y^2)) = \frac{d}{dx}(a) \]
Since \(a\) is a constant, its derivative is \(0\):
\[ \frac{d}{dx}(\tan^{-1}(x^2 + y^2)) = 0 \]

2. **Apply the Chain Rule:**
Using the chain rule, the derivative of \(\tan^{-1}(u)\) where \(u = x^2 + y^2\) is:
\[ \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
Thus,
\[ \frac{1}{1 + (x^2 + y^2)^2} \cdot \frac{d}{dx}(x^2 + y^2) = 0 \]

3. **Differentiate \(u = x^2 + y^2\):**
Now, 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. **Substitute Back:**
Substitute this back into the equation:
\[ \frac{1}{1 + (x^2 + y^2)^2} \cdot (2x + 2y \frac{dy}{dx}) = 0 \]
Since the fraction is not zero, we can set the numerator to zero:
\[ 2x + 2y \frac{dy}{dx} = 0 \]

5. **Solve for \(\frac{dy}{dx}\):**
Rearranging gives:
\[ 2y \frac{dy}{dx} = -2x \]
Dividing both sides by \(2y\) (assuming \(y \neq 0\)):
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{x}{y} \]