The equation of the tangent to the curve x y^ | vedclass

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(B) Given curve: $x y^5+2 x^2 y-x^3+y+1=0$ At $x=0$,the equation becomes $0+0-0+y+1=0$,which gives $y=-1$. So,the point of contact is $(0, -1)$. Diffe

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$y=x-1$

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(B) Given curve: $x y^5+2 x^2 y-x^3+y+1=0$ At $x=0$,the equation becomes $0+0-0+y+1=0$,which gives $y=-1$. So,the point of contact is $(0, -1)$. Differentiating the equation with respect to $x$: $\frac{d}{dx}(x y^5) + \frac{d}{dx}(2 x^2 y) - \frac{d}{dx}(x^3) + \frac{d}{dx}(y) + \frac{d}{dx}(1) = 0$ $y^5 + 5 x y^4 \frac{d y}{d x} + 4 x y + 2 x^2 \frac{d y}{d x} - 3 x^2 + \frac{d y}{d x} = 0$ Substituting $(x, y) = (0, -1)$: $(-1)^5 + 5(0)(-1)^4 \frac{d y}{d x} + 4(0)(-1) + 2(0)^2 \frac{d y}{d x} - 3(0)^2 + \frac{d y}{d x} = 0$ $-1 + 0 + 0 + 0 - 0 + \frac{d y}{d x} = 0$ $\frac{d y}{d x} = 1$ The equation of the tangent at $(0, -1)$ with slope $m=1$ is: $y - (-1) = 1(x - 0)$ $y + 1 = x$ $y = x - 1$

The equation of the tangent to the curve $x y^5+2 x^2 y-x^3+y+1=0$ at $x=0$ is

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