If `y=btan^(-1)(x/a+tan^(-1) `y`/x),f i nd (dy)/(dx)dot`

Explanation

To find \(\frac{dy}{dx}\) for the equation \(y = b \tan^{-1}\left(\frac{x}{a} + \tan^{-1}\left(\frac{y}{x}\right)\right)\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y = b \tan^{-1}\left(\frac{x}{a} + \tan^{-1}\left(\frac{y}{x}\right)\right) \]
Dividing both sides by \(b\):
\[ \frac{y}{b} = \tan^{-1}\left(\frac{x}{a} + \tan^{-1}\left(\frac{y}{x}\right)\right) \]

### Step 2: Applying the Tangent Function
Taking the tangent of both sides:
\[ \tan\left(\frac{y}{b}\right) = \frac{x}{a} + \tan^{-1}\left(\frac{y}{x}\right) \]

### Step 3: Differentiating Both Sides
Now we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}\left(\tan\left(\frac{y}{b}\right)\right) = \frac{d}{dx}\left(\frac{x}{a} + \tan^{-1}\left(\frac{y}{x}\right)\right) \]
Using the chain rule on the left side:
\[ \sec^2\left(\frac{y}{b}\right) \cdot \frac{1}{b} \cdot \frac{dy}{dx} \]
On the right side, we differentiate each term:
\[ \frac{1}{a} + \frac{d}{dx}\left(\tan^{-1}\left(\frac{y}{x}\right)\right) \]

### Step 4: Differentiating \(\tan^{-1}\left(\frac{y}{x}\right)\)
Using the chain rule and quotient rule:
\[ \frac{d}{dx}\left(\tan^{-1}\left(\frac{y}{x}\right)\right) = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{y}{x}\right) \]
Now applying the quotient rule:
\[ \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x \cdot \frac{dy}{dx} - y}{x^2} \]
Thus,
\[ \frac{d}{dx}\left(\tan^{-1}\left(\frac{y}{x}\right)\right) = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{x \cdot \frac{dy}{dx} - y}{x^2} \]

### Step 5: Combining the Derivatives
Now we combine everything:
\[ \frac{1}{b} \sec^2\left(\frac{y}{b}\right) \frac{dy}{dx} = \frac{1}{a} + \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{x \cdot \frac{dy}{dx} - y}{x^2} \]

### Step 6: Isolate \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{1}{b} \sec^2\left(\frac{y}{b}\right) \frac{dy}{dx} - \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{x \cdot \frac{dy}{dx}}{x^2} = \frac{1}{a} - \frac{y}{x^2 + y^2} \]

### Step 7: Solve for \(\frac{dy}{dx}\)
Finally, we can express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\):
\[ \frac{dy}{dx} = \frac{\frac{1}{a} - \frac{y}{x^2 + y^2}}{\frac{1}{b} \sec^2\left(\frac{y}{b}\right) - \frac{x}{x^2 + y^2}} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\frac{1}{a} - \frac{y}{x^2 + y^2}}{\frac{1}{b} \sec^2\left(\frac{y}{b}\right) - \frac{x}{x^2 + y^2}} \]

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