Birinchidan, bizga \(\left( {4, - 2,0} \right)\) da gradient va uning qiymati kerak bo'ladi.

\[\begin{align*}\nabla f & = \left\langle {2{{\bf{e}}^{2x}}\cos \left( {y - 2z} \right), - {{\bf{e}}^{2x}}\sin \left( {y - 2z} \right),2{{\bf{e}}^{2x}}\sin \left( {y - 2z} \right)} \right\rangle \\ \nabla f\left( {4, - 2,0} \right) & = \left\langle {2{{\bf{e}}^8}\cos \left( { - 2} \right), - {{\bf{e}}^8}\sin \left( { - 2} \right),2{{\bf{e}}^8}\sin \left( { - 2} \right)} \right\rangle = \left\langle { - 2481.03,2710.58, - 5421.15} \right\rangle \end{align*}\]

2-qadam

Endi, sinfdagi teorema bo'yicha biz bilamizki, ko'rib chiqilayotgan nuqtadagi maksimal o'zgarish tezligi oddiygina \(\left( {4, - 2,0} \right)\) gradienti bo'lib, biz oldingi bosqichda topdik. Demak, funktsiyaning maksimal o'zgarish tezligi sodir bo'ladigan yo'nalish,

\[\nabla f\left( {4, - 2,0} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle { - 2481.03,2710.58, - 5421.15} \right\rangle }}\]

3-qadam

O'zgarishning maksimal tezligi oddiygina oldingi bosqichdagi gradientning kattaligidir. Shunday qilib, funktsiyaning maksimal o'zgarish tezligi,

\[\left\| {\nabla f\left( {4, - 2,0} \right)} \right\| = \sqrt {{{\left( { - 2481.03} \right)}^2} + {{\left( {2710.58} \right)}^2} + {{\left( { - 5421.15} \right)}^2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6549.17}}\]