log (x2 + y2) = 2 tan-1 (y/x). Find dy/dx?

1.	log (x2 + y2) = 2 tan-1 (y/x). Find dy/dx?
Answer» \(log(x^2 + y^2) = 2tan^{-1} \left(\frac yx\right)\) Differentiate both sides w.r.t. x, we get \(\therefore \frac1{x^2 + y^2} \left(2x + 2y \frac{dy}{dx}\right) = \frac2{1 + \left(\frac yx\right)^2} \left(\frac{x \frac{dy}{dx} - y}{x^2}\right)\) ⇒ \(\frac{x + y \frac{dy}{dx}}{x^2 + y^2} = \frac{x^2}{x^2 + y^2 }\;\frac{\left(x\frac{dy}{dx} - y\right)}{x^2}\) ⇒ \(x + y \,\frac{dy}{dx} = x\,\frac{dy}{dx} - y\) ⇒ \((y - x) \frac{dy}{dx} = -(x + y)\) ⇒ \(\frac{dy}{dx} = \frac{-(x+y)}{y - x} = \frac{x + y}{x - y}\)

Answer»

\(log(x^2 + y^2) = 2tan^{-1} \left(\frac yx\right)\)

Differentiate both sides w.r.t. x, we get

\(\therefore \frac1{x^2 + y^2} \left(2x + 2y \frac{dy}{dx}\right) = \frac2{1 + \left(\frac yx\right)^2} \left(\frac{x \frac{dy}{dx} - y}{x^2}\right)\)

⇒ \(\frac{x + y \frac{dy}{dx}}{x^2 + y^2} = \frac{x^2}{x^2 + y^2 }\;\frac{\left(x\frac{dy}{dx} - y\right)}{x^2}\)

⇒ \(x + y \,\frac{dy}{dx} = x\,\frac{dy}{dx} - y\)

⇒ \((y - x) \frac{dy}{dx} = -(x + y)\)

⇒ \(\frac{dy}{dx} = \frac{-(x+y)}{y - x} = \frac{x + y}{x - y}\)

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