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Supongo que pides la pendiente de la tangente a la curva en el punto (3,1)
(1) derivemos de forma implícita haciendo![y = y(x) y = y(x)](https://tex.z-dn.net/?f=y+%3D+y%28x%29)
![\dfrac{d}{dx}[3(x^2+y^2)^2]=\dfrac{d}{dx}(100xy)\\ \\ \\
3\cdot\dfrac{d}{dx}[(x^2+y^2)^2]=100\cdot\dfrac{d}{dx}(xy)\\ \\ \\
6(x^2+y^2)\cdot\dfrac{d}{dx}(x^2+y^2)=100\cdot \left(\dfrac{dx}{dx}\cdot y+x\cdot \dfrac{dy}{dx}\right)\\ \\ \\
3(x^2+y^2)\cdot\left(\dfrac{d(x^2)}{dx}+\dfrac{d(y^2)}{dx}\right)=50\cdot \left(y+x\cdot \dfrac{dy}{dx}\right)
\dfrac{d}{dx}[3(x^2+y^2)^2]=\dfrac{d}{dx}(100xy)\\ \\ \\
3\cdot\dfrac{d}{dx}[(x^2+y^2)^2]=100\cdot\dfrac{d}{dx}(xy)\\ \\ \\
6(x^2+y^2)\cdot\dfrac{d}{dx}(x^2+y^2)=100\cdot \left(\dfrac{dx}{dx}\cdot y+x\cdot \dfrac{dy}{dx}\right)\\ \\ \\
3(x^2+y^2)\cdot\left(\dfrac{d(x^2)}{dx}+\dfrac{d(y^2)}{dx}\right)=50\cdot \left(y+x\cdot \dfrac{dy}{dx}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdx%7D%5B3%28x%5E2%2By%5E2%29%5E2%5D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%28100xy%29%5C%5C+%5C%5C+%5C%5C%0A3%5Ccdot%5Cdfrac%7Bd%7D%7Bdx%7D%5B%28x%5E2%2By%5E2%29%5E2%5D%3D100%5Ccdot%5Cdfrac%7Bd%7D%7Bdx%7D%28xy%29%5C%5C+%5C%5C+%5C%5C%0A6%28x%5E2%2By%5E2%29%5Ccdot%5Cdfrac%7Bd%7D%7Bdx%7D%28x%5E2%2By%5E2%29%3D100%5Ccdot+%5Cleft%28%5Cdfrac%7Bdx%7D%7Bdx%7D%5Ccdot+y%2Bx%5Ccdot+%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A3%28x%5E2%2By%5E2%29%5Ccdot%5Cleft%28%5Cdfrac%7Bd%28x%5E2%29%7D%7Bdx%7D%2B%5Cdfrac%7Bd%28y%5E2%29%7D%7Bdx%7D%5Cright%29%3D50%5Ccdot+%5Cleft%28y%2Bx%5Ccdot+%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%0A)
![3(x^2+y^2)\cdot\left(2x+2y\cdot\dfrac{dy}{dx}\right)=50\cdot \left(y+x\cdot \dfrac{dy}{dx}\right)\\ \\ \\
3(x^2+y^2)\cdot\left(x+y\cdot\dfrac{dy}{dx}\right)=25\cdot \left(y+x\cdot \dfrac{dy}{dx}\right)
3(x^2+y^2)\cdot\left(2x+2y\cdot\dfrac{dy}{dx}\right)=50\cdot \left(y+x\cdot \dfrac{dy}{dx}\right)\\ \\ \\
3(x^2+y^2)\cdot\left(x+y\cdot\dfrac{dy}{dx}\right)=25\cdot \left(y+x\cdot \dfrac{dy}{dx}\right)](https://tex.z-dn.net/?f=3%28x%5E2%2By%5E2%29%5Ccdot%5Cleft%282x%2B2y%5Ccdot%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%3D50%5Ccdot+%5Cleft%28y%2Bx%5Ccdot+%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A3%28x%5E2%2By%5E2%29%5Ccdot%5Cleft%28x%2By%5Ccdot%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%3D25%5Ccdot+%5Cleft%28y%2Bx%5Ccdot+%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%29%0A)
(2) Hallemos la pendiente de la tangente en el punto (3,1)
![3(3^2+1^2)\cdot\left(3+1\cdot\left.\dfrac{dy}{dx}\right|_{(x,y)=(3,1)}\right)=25\cdot \left(1+3\cdot \left.\dfrac{dy}{dx}\right|_{(x,y)=(3,1)}\right)\\ \\ \\
30(3+m)=25(1+3m)\\ \\ \\
\boxed{m= \dfrac{13}{9}} 3(3^2+1^2)\cdot\left(3+1\cdot\left.\dfrac{dy}{dx}\right|_{(x,y)=(3,1)}\right)=25\cdot \left(1+3\cdot \left.\dfrac{dy}{dx}\right|_{(x,y)=(3,1)}\right)\\ \\ \\
30(3+m)=25(1+3m)\\ \\ \\
\boxed{m= \dfrac{13}{9}}](https://tex.z-dn.net/?f=3%283%5E2%2B1%5E2%29%5Ccdot%5Cleft%283%2B1%5Ccdot%5Cleft.%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%7C_%7B%28x%2Cy%29%3D%283%2C1%29%7D%5Cright%29%3D25%5Ccdot+%5Cleft%281%2B3%5Ccdot+%5Cleft.%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%7C_%7B%28x%2Cy%29%3D%283%2C1%29%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0A30%283%2Bm%29%3D25%281%2B3m%29%5C%5C+%5C%5C+%5C%5C%0A%5Cboxed%7Bm%3D+%5Cdfrac%7B13%7D%7B9%7D%7D)
(1) derivemos de forma implícita haciendo
(2) Hallemos la pendiente de la tangente en el punto (3,1)
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