Analiza matematyczna, zadanie nr 1275
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Zadanie / Rozwi膮zanie |
gam post贸w: 1 |  2013-04-18 17:36:38witam jak policzy膰 pochodn膮 drugiego rz臋du z pochodnej a) y(ln(x^{2}+y^{2})+\frac{2x^{2}}{x^{2}+y^{2}}) b) x(ln(x^{2}+y^{2})+\frac{2y^{2}}{x+y^{2}}) z g贸ry wielkie dzi臋ki ;d |
abcdefgh post贸w: 1255 | 2013-04-24 21:26:20 $ y(ln(x^{2}+y^{2})+\frac{2x^{2}}{x^{2}+y^{2}})$ a) f\'(x)=$ y[(\frac{1}{x^2+y^2}*2x)+\frac{4x*(x^2+y^2)-2x*2x^2}{(x^2+y^2)^2}$=$y(\frac{2x}{x^2+y^2}+\frac{4x^3+4xy^2-4x^3}{(x^2+y^2)^2}$=$y(\frac{2x(x^2+y^2)+4xy^2}{(x^2+y^2)^2})$ =$\frac{y(2x^3+2xy^2+4xy^2)}{(x^2+y^2)^2}$=$\frac{2y(x^3+3xy^2)}{(x^2+y^2)^2}$ f\"(x)=$\frac{6x^2y(x^2+y^2)^2-2yx^3*2(x^2+y^2)*2x+6y^3(x^2+y^2)^2-6xy^3*2(x^2+y^2)*2x}{(x^2+y^2)^4}$=$\frac{(x^2+y^2)*[6x^2y(x^2+y^2)-8x^4y+6y^3(x^2+y^2)-24x^2y^3]}{(x^2+y^2)^4}$ =$\frac{6x^4y+6x^2y^3-8x^4y+6x^2y^3+6y^5-24x^2y^3}{(x^2+y^2)^3}$=$\frac{-2x^4y-12x^2y^3+6y^5}{(x^2+y^2)^3}$=$\frac{2y(-x^4-6x^2y^2+3y^4)}{(x^2+y^2)^3}$ Wiadomo艣膰 by艂a modyfikowana 2013-04-25 21:35:54 przez abcdefgh |
abcdefgh post贸w: 1255 | 2013-04-25 22:23:54 $ a) y(ln(x^{2}+y^{2})+\frac{2x^{2}}{x^{2}+y^{2}})$ $f\'(y)=y*ln(x^2+y^2)+\frac{2x^2y}{x^2+y^2}$= $ln(x^2+y^2)*1+y*\frac{1}{x^2+y^2}*2y+\frac{2x^2(x^2+y^2)-2x^2y*2y}{(x^2+y^2)^2}$=$ln(x^2+y^2)+\frac{2y^2(x^2+y^2)+(2x^4+2x^2y^2-4x^2y^2)}{(x^2+y^2)^2}$= $ln(x^2+y^2)+\frac{2x^2y^2+2y^4+2x^4-2x^2y^2}{(x^2+y^2)^2}$ =$ln(x^2+y^2)+\frac{2(x^4+y^4)}{(x^2+y^2)^2}$ f\"(y)=$\frac{1}{x^2+y^2}*2y+\frac{0-2x^4*2(x^2+y^2)*2y+8y^3*(x^2+y^2)^2-2y^4*2(x^2+y^2)*2y}{(x^2+y^2)^4}$=$ \frac{(x^2+y^2)[-8x^4y+8y^3(x^2+y^2)-8y^5]}{(x^2+y^2)^4}$= $\frac{-8x^4y+8x^2y^3+8y^5-8y^5}{(x^2+y^2)^3}$=$ \frac{-8x^4y+8x^2y^3}{(x^2+y^2)^3}$=$ \frac{8x^2y(-x^2+y^2)}{(x^2+y^2)^3}$ |
abcdefgh post贸w: 1255 | 2013-04-25 22:48:27 $ b) x(ln(x^{2}+y^{2})+\frac{2y^{2}}{x+y^{2}})$ $f\'(x)=ln(x^2+y^2)*1+\frac{x}{x^2+y^2}*2x+\frac{0-2y^2*1}{(x+y^2)^2}$=$ln(x^2+y^2)+\frac{2x^2}{x^2+y^2}+\frac{-2y^2}{(x+y^2)^2}$ f\"(x)=$\frac{1}{x^2+y^2}*2x+\frac{4x*(x^2+y^2)-2x^2*2x}{(x^2+y^2)^2}+\frac{0+2y^2*2(x+y^2)*1}{(x+y^2)^4}$=$ \frac{2x}{x^2+y^2}+\frac{4xy^2}{(x^2+y^2)^2}+\frac{4y^2(x+y^2)}{(x+y^2)^4}$ $f\'(y)=0+\frac{x}{x^2+y^2}*2y+\frac{4yx(x+y^2)-2y^2x*2y}{(x+y^2)^2}$=$\frac{2yx}{x^2+y^2}+\frac{4x^2y+4xy^3-4y^3x}{(x+y^2)^2}$=$\frac{2yx}{x^2+y^2}+\frac{4x^2y}{(x+y^2)^2}$=$ $f\"(y)=\frac{2y(x^2+y^2)-2yx*2y}{(x^2+y^2)^2}+\frac{4x^2(x+y^2)^2-4x^2y*2(x+y^2)*2y}{(x+y^2)^4}$=$ \frac{2x^2y+2y^3-4y^2x}{(x^2+y^2)^2}+\frac{4x^3+4x^2y^2-16x^2y^2(x+y^2)}{(x+y^2)^4}$ |
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