Question

The slope of the line tangent to the curve 2^x + 3x^2y + y = 1 at the point (0, 1) is?

Answer 1

`\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}}`


`\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}`