Respuesta :

[tex]\bf 2^x+3x^2y+y=1\\\\ -----------------------------\\\\ 2^xln(2)+3\left( 2xy+x^2\cfrac{dy}{dx} \right)+\cfrac{dy}{dx}=0 \\\\\\ 3x^2\cfrac{dy}{dx}+\cfrac{dy}{dx}=-2^xln(2)-6xy \\\\\\ \cfrac{dy}{dx}(3x^2+1)=-2^xln(2)-6xy \\\\\\ \boxed{\cfrac{dy}{dx}=\cfrac{-2^xln(2)-6xy}{3x^2+1}}[/tex]

[tex]\bf \left. \cfrac{dy}{dx}=\cfrac{-2^xln(2)-6xy}{3x^2+1} \right|_{0,1}\implies \cfrac{dy}{dx}=\cfrac{-2^0ln(2)-6(0)(1)}{3(0)^2+1} \\\\\\ \cfrac{dy}{dx}=-ln(2)\\\\ -----------------------------\\\\ y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-1=-ln(2)(x-0)\\ \left. \qquad \right.\uparrow\\ \textit{point-slope form} \\\\\\ \boxed{y=-ln(2)x+1}[/tex]