Question

achar derivada y'=dy/dx da seguinte funcao implicita y: xy=arctgx/y

Answer 1

$xy=\mathrm{arctg}\!\left(\dfrac{x}{y} \right )$


Sabendo que $y$ é função de $x,$ derivando os dois lados, temos

$\dfrac{d}{dx}(xy)=\dfrac{d}{dx}\!\left[\mathrm{arctg}\!\left(\dfrac{x}{y} \right ) \right ]\\\\\\ \dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}=\dfrac{1}{1+\left(\frac{x}{y}\right)^{\!\!2}}\cdot \dfrac{d}{dx}\!\left(\frac{x}{y} \right )\\\\\\ 1\cdot y+x\cdot \dfrac{dy}{dx}=\dfrac{1}{1+\frac{x^2}{y^2}}\cdot \dfrac{\frac{d}{dx}(x)\cdot y-x\cdot \frac{dy}{dx}}{y^2}\\\\\\ y+x\cdot \dfrac{dy}{dx}=\dfrac{1}{1+\frac{x^2}{y^2}}\cdot \dfrac{1\cdot y-x\cdot \frac{dy}{dx}}{y^2}$

$y+x\cdot \dfrac{dy}{dx}=\dfrac{y-x\cdot\frac{dy}{dx}}{\left(1+\frac{x^2}{y^2} \right )y^2}\\\\\\ y+x\cdot \dfrac{dy}{dx}=\dfrac{y-x\cdot\frac{dy}{dx}}{y^2+x^2}\\\\\\ (x^2+y^2)\cdot \left(y+x\cdot \dfrac{dy}{dx} \right )=y-x\cdot\dfrac{dy}{dx}\\\\\\ (x^2+y^2)\,y+(x^2+y^2)\,x\cdot \dfrac{dy}{dx}=y-x\cdot\dfrac{dy}{dx}\\\\\\ (x^2+y^2)\,x\cdot \dfrac{dy}{dx}+x\cdot\dfrac{dy}{dx}=y-(x^2+y^2)\,y$

$\left[(x^2+y^2)\,x+x \right ]\cdot \dfrac{dy}{dx}=y-(x^2+y^2)\,y\\\\\\ \dfrac{dy}{dx}=\dfrac{y-(x^2+y^2)\,y}{(x^2+y^2)\,x+x}\\\\\\ \dfrac{dy}{dx}=\dfrac{y\,[1-(x^2+y^2)]}{x\,[(x^2+y^2)+1]}\\\\\\ \therefore~~\boxed{\begin{array}{c} \dfrac{dy}{dx}=\dfrac{y\,(1-x^2-y^2)}{x\,(1+x^2+y^2)} \end{array}}$