13.7-bo'lim : Yo'nalishli hosilalar
1 – 4 masalalar uchun berilgan funksiyaning gradientini aniqlang.
- \(f\left( {x,y} \right) = {y^3}{x^5} + \ln \left( {xy} \right)\)
- \(f\left( {x,y} \right) = {{\bf{e}}^{\frac{x}{y}}} + {y^4}\sin \left( {xy} \right)\)
- \(\displaystyle f\left( {x,y,z} \right) = 4z - \frac{{{y^4}}}{{2{z^3}}} + \sqrt {{x^3}} \left( {z - 1} \right)\)
- \(\displaystyle f\left( {x,y,z} \right) = \cos \left( {\frac{{xy}}{z}} \right) + {z^3}{y^2}x\)
5-8-masalalar uchun ko'rsatilgan yo'nalishda berilgan funktsiya uchun \({D_{\vec u}}f\) ni aniqlang.
- \(f\left( {x,y} \right) = \ln \left( {2xy} \right) - \sin \left( {{x^2} + {y^2}} \right)\) \(\vec v = \left\langle {7, - 3} \right\rangle \) yo'nalishida
- \(f\left( {x,y} \right) = 4{x^2}{y^3} - \sqrt {2x + 5y} \) \(\vec v = \left\langle { - 1,4} \right\rangle \) yo'nalishida
- \(\displaystyle f\left( {x,y,z} \right) = 8x{y^2} - \frac{{5{z^2}}}{x} + {y^4}\) \(\vec v = \left\langle { - 4,1,2} \right\rangle \) yo'nalishida
- \(\displaystyle f\left( {x,y,z} \right) = \frac{{3x}}{{{y^2} - {z^3}}} + 5{x^2} - 8y\) \(\vec v = \left\langle {0,3, - 2} \right\rangle \) yo'nalishida
- \(\vec v = \left\langle {2, - 3,6} \right\rangle \) ning \(f\left( {x,y,z} \right) = {{\bf{e}}^{x\,y{\,^2}}} + 4z{y^3}\) yo‘nalishi uchun \({D_{\vec u}}f\left( { - 1,4,6} \right)\) ni aniqlang.
- \(\vec v = \left\langle {1,5,2} \right\rangle \) ning \(\displaystyle f\left( {x,y,z} \right) = \ln \left( {\frac{x}{z}} \right) + \ln \left( {\frac{z}{y}} \right) + {y^2}x\) yo‘nalishi uchun \({D_{\vec u}}f\left( {8,1,2} \right)\) ni aniqlang.
11-13-masalalar uchun funksiyaning ko'rsatilgan nuqtadagi maksimal o'zgarish tezligini va bu maksimal o'zgarish tezligi sodir bo'lgan yo'nalishni toping.
- \(f\left( {x,y} \right) = {{\bf{e}}^{4x\,y}}\), \(\left( {6, - 2} \right)\)
- \(f\left( {x,y,z} \right) = {x^2}{y^4} - 3{z^2}x\), \(\left( {1, - 6,3} \right)\)
- \(\displaystyle f\left( {x,y,z} \right) = \ln \left( {\frac{{2x + 3y}}{z}} \right)\), \(\left( {2,7,4} \right)\)
- Berilgan \(\displaystyle \vec u = \left\langle { - \frac{3}{5}, - \frac{4}{5}} \right\rangle \), \(\displaystyle \vec v = \left\langle {\frac{4}{{\sqrt {20} }},\frac{2}{{\sqrt {20} }}} \right\rangle \), \(\displaystyle \vec w = \left\langle { - \frac{3}{{\sqrt {11} }}, - \frac{2}{{\sqrt {11} }}} \right\rangle \), \(\displaystyle {D_{\vec u}}\left( { - 1,4} \right) = \frac{{14}}{5}\) va \(\displaystyle {D_{\vec v}}\left( { - 1,4} \right) = - \frac{{22}}{{\sqrt {20} }}\) \({D_{\vec w}}\left( { - 1,4} \right)\) qiymatini aniqlaydi.
- Berilgan \(\displaystyle \vec u = \left\langle {\frac{1}{{\sqrt {15} }},\frac{4}{{\sqrt {15} }}} \right\rangle \), \(\displaystyle \vec v = \left\langle { - \frac{3}{{\sqrt {34} }}, - \frac{5}{{\sqrt {34} }}} \right\rangle \), \(\displaystyle \vec w = \left\langle { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right\rangle \), \(\displaystyle {D_{\vec u}}\left( {0,1} \right) = \frac{{18}}{{\sqrt {15} }}\) va \(\displaystyle {D_{\vec v}}\left( {0,1} \right) = - \frac{{40}}{{\sqrt {34} }}\) \({D_{\vec w}}\left( {0,1} \right)\) qiymatini aniqlaydi.