Esecrevendo a equação da reta tangente à curva y2 - x4 = 3 que passa pelo ponto (1,2) temos:
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Acompanhe a derivação implícita:
![\displaystyle y^{2}-x^{4}=3 \\ \\ \\ 2y \, \frac{dy}{dx} - 4x^{3}=0 \\ \\ \\ 2y \, \frac{dy}{dx} = 4x^{3} \\ \\ \\ \frac{dy}{dx} = \frac{4x^{3}}{2y} \\ \\ \\ \frac{dy}{dx} = \frac{2x^{3}}{y} \displaystyle y^{2}-x^{4}=3 \\ \\ \\ 2y \, \frac{dy}{dx} - 4x^{3}=0 \\ \\ \\ 2y \, \frac{dy}{dx} = 4x^{3} \\ \\ \\ \frac{dy}{dx} = \frac{4x^{3}}{2y} \\ \\ \\ \frac{dy}{dx} = \frac{2x^{3}}{y}](https://tex.z-dn.net/?f=%5Cdisplaystyle+y%5E%7B2%7D-x%5E%7B4%7D%3D3+%5C%5C+%5C%5C+%5C%5C+2y+%5C%2C+%5Cfrac%7Bdy%7D%7Bdx%7D+-+4x%5E%7B3%7D%3D0+%5C%5C+%5C%5C+%5C%5C+2y+%5C%2C+%5Cfrac%7Bdy%7D%7Bdx%7D+%3D+4x%5E%7B3%7D+%5C%5C+%5C%5C+%5C%5C+%5Cfrac%7Bdy%7D%7Bdx%7D+%3D+%5Cfrac%7B4x%5E%7B3%7D%7D%7B2y%7D+%5C%5C+%5C%5C+%5C%5C+%5Cfrac%7Bdy%7D%7Bdx%7D+%3D+%5Cfrac%7B2x%5E%7B3%7D%7D%7By%7D)
A inclinação m será:
![\displaystyle m = \frac{2x^{3}}{y} \\ \\ \\ m = \frac{2 \cdot 1^{3}}{2} \\ \\ \\ m = 1 \displaystyle m = \frac{2x^{3}}{y} \\ \\ \\ m = \frac{2 \cdot 1^{3}}{2} \\ \\ \\ m = 1](https://tex.z-dn.net/?f=%5Cdisplaystyle+m+%3D+%5Cfrac%7B2x%5E%7B3%7D%7D%7By%7D+%5C%5C+%5C%5C+%5C%5C+m+%3D+%5Cfrac%7B2+%5Ccdot+1%5E%7B3%7D%7D%7B2%7D+%5C%5C+%5C%5C+%5C%5C+m+%3D+1)
E a equação geral da reta tangente à curva é:
![y-y_{0}=m(x-x_{0}) \\ \\ \\ y-2=1(x-1) \\ \\ \\ y-2=x-1 \\ \\ \\ y=x-1+2 \\ \\ \\ y=x+1 \\ \\ \\ \boxed{\boxed{r: \, -x+y-1=0}} y-y_{0}=m(x-x_{0}) \\ \\ \\ y-2=1(x-1) \\ \\ \\ y-2=x-1 \\ \\ \\ y=x-1+2 \\ \\ \\ y=x+1 \\ \\ \\ \boxed{\boxed{r: \, -x+y-1=0}}](https://tex.z-dn.net/?f=y-y_%7B0%7D%3Dm%28x-x_%7B0%7D%29+%5C%5C+%5C%5C+%5C%5C+y-2%3D1%28x-1%29+%5C%5C+%5C%5C+%5C%5C+y-2%3Dx-1+%5C%5C+%5C%5C+%5C%5C+y%3Dx-1%2B2+%5C%5C+%5C%5C+%5C%5C+y%3Dx%2B1+%5C%5C+%5C%5C+%5C%5C+%5Cboxed%7B%5Cboxed%7Br%3A+%5C%2C+-x%2By-1%3D0%7D%7D+)
A inclinação m será:
E a equação geral da reta tangente à curva é:
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