4.12-bo'lim : Differensiallar
4. \(y = {{\bf{e}}^{{x^{\,2}}}}\) uchun \(dy\) va \(\Delta y\) ni hisoblang, chunki \(x\) 3 dan 3.01 ga o'zgaradi.
Yechim
Birinchidan, \(\Delta y\)-ning haqiqiy o'zgarishini olaylik.
\[\Delta y = {{\bf{e}}^{3.01{\,^2}}} - \,{{\bf{e}}^{3{\,^2}}} = 501.927\]2-qadam
Keyinchalik, bizga differentsial kerak bo'ladi.
\[dy = 2x\,{{\bf{e}}^{{x^{\,2}}}}dx\]3-qadam
\(x\) 3 dan 3.01 ga o'zgarganda bizda \(\Delta x = 3.01 - 3 = 0.01\) bor va biz \(dx \approx \Delta x = 0.01\) deb faraz qilamiz. Taxminiy o'zgarish, \(dy\), keyin,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{dy = 2\left( 3 \right)\,{{\bf{e}}^{{3^{\,2}}}}\left( {0.01} \right) = 486.185}}\]\(x\) ning "boshlang'ich" qiymatidan foydalanishni unutmang (i.e. \(x = 3\)) differensialdagi barcha \(x\)lar uchun.