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Implicit function help (1 Viewer)
- Thread starter 08er
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Count Chocular
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- 2012
wat he sed}{\mathrm{d} x}\ \ =\ \ \frac{\mathrm{d} 13}{\mathrm{d} x}\\ \\\therefore \ \ \ \frac{\mathrm{d} y^{3}}{\mathrm{d} y}\ \cdot \frac{\mathrm{d} y}{\mathrm{d} x}\ \ \ +\ \ \ 3\ \left \{x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}\ + \ y\frac{\mathrm{d} x^{2}}{\mathrm{d} x} \right \}\ \ =\ \ 0\\ \\\therefore \ \ \ 3y^{2}.\frac{\mathrm{d} y}{\mathrm{d} x}\ \ +\ \ 3\left \{x^{2}\frac{\mathrm{d} y}{\mathrm{d} x}\ + \ 2xy \right \}\ \ = \ 0\\ \\ \therefore \ \ \ 3\left (x^{2}\ +\ y^{2} \right )\frac{\mathrm{d} y}{\mathrm{d} x}\ \ \ =\ \ \ -6xy\\ \\ \therefore \ \ \ \frac{\mathrm{d} y}{\mathrm{d} x}\ \ \ =\ \ \ \frac{-2xy}{x^{2}\ +\ y^{2}})
It was. I whacked out the 3 and I held x^2 x the derivative of y (which is dy/dx) + I held y x the derivative of x^2 (which is 2x).
Isn't this 4 unit? Implicit functions?
clintmyster
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i think it is tho you could use it in some 3u work.