Question

Calcule la derivada implícita de la Siguiente función ln⁡(xy)-e^(x/y)=0

Answer 1

$\text{Derivemos...}\\ \\\dfrac{d}{dx}\left(\ln(xy)-e^{x/y}\right)=0\\ \\ \\\dfrac{d}{dx}\ln(xy)-\dfrac{d}{dx}e^{x/y}=0\\ \\ \\\dfrac{\dfrac{d}{dx}(xy)}{xy}-\dfrac{d}{dx}(\dfrac{x}{y})\cdot e^{x/y}=0\\ \\\\\dfrac{y\dfrac{dx}{dx}+x\dfrac{dy}{dx}}{xy}-\dfrac{y\dfrac{dx}{dx}-x\dfrac{dy}{dx}}{y^2}\cdot e^{x/y}=0\\ \\ \\\dfrac{y+xy'}{xy}-\dfrac{y-xy'}{y^2}\cdot e^{x/y}=0\\ \\ \\\dfrac{1}{x}+\dfrac{y'}{y}-\dfrac{1}{y}+\dfrac{xy'}{y^2}\cdot e^{x/y}=0$

$y'(\dfrac{1}{y}+\dfrac{xe^{x/y}}{y^2})+(\dfrac{1}{x}-\dfrac{1}{y})=0\\ \\ \\ \\y'(\dfrac{1}{y}+\dfrac{xe^{x/y}}{y^2})=\dfrac{1}{y}-\dfrac{1}{x}\\ \\ \\y'=\dfrac{\dfrac{1}{y}-\dfrac{1}{x}}{\dfrac{1}{y}+\dfrac{xe^{x/y}}{y^2}}\\ \\ \\\\y'=\dfrac{(x-y)y}{x(y+xe^{1/x})}$