4.12-bo'lim : Differensiallar
5. \(y = {x^5} - 2{x^3} + 7x\) uchun \(dy\) va \(\Delta y\) ni hisoblang, chunki \(x\) 6 dan 5,9 ga o'zgaradi.
Yechim
Birinchidan, \(\Delta y\)-ning haqiqiy o'zgarishini olaylik.
\[\Delta y = \left( {{{5.9}^5} - 2\left( {{{5.9}^3}} \right) + 7\left( {5.9} \right)} \right) - \left( {{6^5} - 2\left( {{6^3}} \right) + 7\left( 6 \right)} \right) = - 606.215\]2-qadam
Keyinchalik, bizga differentsial kerak bo'ladi.
\[dy = \left( {5{x^4} - 6{x^2} + 7} \right)dx\]3-qadam
\(x\) 6 dan 5,9 ga o‘zgarganda bizda \(\Delta x = 5.9 - 6 = - 0.1\) bor va biz \(dx \approx \Delta x = - 0.1\) deb faraz qilamiz. Taxminiy o'zgarish, \(dy\), keyin,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{dy = \left( {5\left( {{6^4}} \right) - 6\left( {{6^2}} \right) + 7} \right)\left( { - 0.1} \right) = - 627.1}}\]\(x\) ning "boshlang'ich" qiymatidan foydalanishni unutmang (i.e. \(x = 6\)) differensialdagi barcha \(x\)lar uchun.