Question

Derivar implicitamente xy = cotg(xy)

Answer 1

$xy=\mathrm{cotg}(xy)$


Derivando implicitamente ambos os lados, em relação a $x,$

$\dfrac{d}{dx}(xy)=\dfrac{d}{dx}[\mathrm{cotg}(xy)]\\ \\ \\ \dfrac{d}{dx}(xy)=-\mathrm{cotg}(xy)\cdot \mathrm{cossec}(xy)\cdot \dfrac{d}{dx}(xy)\\ \\ \\ \dfrac{d}{dx}(xy)+\mathrm{cotg}(xy)\cdot \mathrm{cossec}(xy)\cdot \dfrac{d}{dx}(xy)=0$


Colocando $\dfrac{d}{dx}(xy)$ em evidência no lado esquerdo, temos

$\dfrac{d}{dx}(xy)\left[1+\mathrm{cotg}(xy)\cdot \mathrm{cossec}(xy) \right ]=0\\ \\ \\ \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{d}{dx}(y) \right ]\cdot \left[1+\mathrm{cotg}(xy)\cdot \mathrm{cossec}(xy) \right ]=0\\ \\ \\ \left[1\cdot y+x\cdot \dfrac{dy}{dx} \right ]\cdot \left[1+\mathrm{cotg}(xy)\cdot \mathrm{cossec}(xy) \right ]=0\\ \\ \\ \left[y+x\,\dfrac{dy}{dx} \right ]\cdot \left[1+\mathrm{cotg}(xy)\cdot \mathrm{cossec}(xy) \right ]=0$