Find `dy/dx" when "tan^(-1)(x^(2)+y^(2))=0`.

Answer

To find \(\frac{dy}{dx}\) when \(\tan^{-1}(x^2 + y^2) = 0\), we can follow these steps:

### Step 1: Simplify the equation
Given:
\[ \tan^{-1}(x^2 + y^2) = 0 \]
This implies:
\[ x^2 + y^2 = \tan(0) \]
Since \(\tan(0) = 0\), we have:
\[ x^2 + y^2 = 0 \]

### Step 2: Rearranging the equation
From the equation \(x^2 + y^2 = 0\), we can express \(y^2\) in terms of \(x^2\):
\[ y^2 = -x^2 \]

### Step 3: Differentiate both sides with respect to \(x\)
Now, we differentiate both sides of the equation \(y^2 = -x^2\) with respect to \(x\):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(-x^2) \]
Using the chain rule on the left side, we get:
\[ 2y \frac{dy}{dx} = -2x \]

### Step 4: Solve for \(\frac{dy}{dx}\)
Now, we can isolate \(\frac{dy}{dx}\):
\[ 2y \frac{dy}{dx} = -2x \]
Dividing both sides by \(2y\) (assuming \(y \neq 0\)):
\[ \frac{dy}{dx} = -\frac{x}{y} \]

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

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