Question

If \(x^2 \tan^{-1}\left(\frac{y}{x}\right) - y^2 \tan^{-1}\left(\frac{x}{y}\right) = k\), then \(\left(\frac{dy}{dx}\right)_{(1,1)} =\) ?

• A. \(0\)
• B. \(\frac{\pi}{4}\)
• C. \(1\)
• D. \(\frac{\pi}{2}\)

Hint:

Use implicit differentiation when a function involving both \(x\) and \(y\) is given as a constant. Evaluate derivatives carefully at the specified point.

Answer 1

The Correct Option is C

Let \(F(x, y) = x^2 \tan^{-1}\left(\frac{y}{x}\right) - y^2 \tan^{-1}\left(\frac{x}{y}\right)\). We are given \(F(x, y) = k\), a constant. So differentiate implicitly: \[ \frac{dF}{dx} = 0 = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} \] Compute partial derivatives at \((x,y) = (1,1)\): \[ \frac{\partial F}{\partial x} = 2x \tan^{-1}\left(\frac{y}{x}\right) + x^2 \cdot \left[\frac{-y}{x^2 + y^2}\right] - 0 - y^2 \cdot \left[\frac{y}{x^2 + y^2}\right] \] \[ \frac{\partial F}{\partial y} = x^2 \cdot \left[\frac{x}{x^2 + y^2}\right] - 2y \tan^{-1}\left(\frac{x}{y}\right) + y^2 \cdot \left[\frac{-x}{x^2 + y^2}\right] \] Plugging in \(x=1, y=1\) and simplifying: \[ \frac{\partial F}{\partial x} = 2 \cdot \frac{\pi}{4} - \frac{1}{2} - \frac{1}{2} = \frac{\pi}{2} - 1 \] \[ \frac{\partial F}{\partial y} = \frac{1}{2} - 2 \cdot \frac{\pi}{4} - \frac{1}{2} = -\frac{\pi}{2} \] Now: \[ 0 = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y} \cdot \frac{dy}{dx} \Rightarrow 0 = \left(\frac{\pi}{2} - 1\right) + (-\frac{\pi}{2}) \cdot \frac{dy}{dx} \] Solving: \(\frac{dy}{dx} = 1\)