Question

Determine a expressão da derivada y=dy/dx para função y=f(x) dada implicitamente pela expressão y+ln(x^(2)+y^(2))=4

Answer 1

$y+ ln(x^2+y^2)=4\\\\ \frac{d}{dx} ( y+ ln(x^2+y^2) )=\frac{d}{dx}(4)\\\\ \frac{dy}{dx}+ \frac{1}{x^2+y^2}*\frac{d}{dx} (x^2+y^2) = 0\\\\ \frac{dy}{dx} + \frac{1}{x^2+y^2} *(2x \; \frac{dx}{dx} + 2y\;\frac{dy}{dx} )=0 \\\\ \frac{dy}{dx} + \frac{1}{x^2+y^2} *(2x + 2y\;\frac{dy}{dx} )=0\\\\ \frac{dy}{dx} + \frac{2x}{x^2+y^2 }+ \frac{2y\frac{dy}{dx}}{x^2+y^2}=0\\\\ \frac{dy}{dx}+ \frac{2y\frac{dy}{dx}}{x^2+y^2}=- \frac{2x}{x^2+y^2} \\\\ \frac{dy}{dx} ( 1+ \frac{2y}{x^2+y^2} )= - \frac{2x}{x^2+y^2}$

$\frac{dy}{dx}= - \frac{2x}{x^2+y^2} * \frac{1}{( 1+ \frac{2y}{x^2+y^2} )} \\\\ \frac{dy}{dx}= -\frac{2x}{x^2+y^2} * \frac{1}{ \frac{x^2+y^2+2y}{x^2+y^2} } \\\\ \frac{dy}{dx}=- \frac{2x}{x^2+y^2}* \frac{x^2+y^2}{x^2+y^2+2y} \\\\ \boxed{\boxed{\frac{dy}{dx}= \frac{-2x}{x^2+y^2+2y} }}$