3.1-bo'lim : Hosilaning ta'rifi
10. hosila ta’rifidan foydalanib, hosilasini toping,
\[Z\left( t \right) = \sqrt {3t - 4} \]Yechim
Avval biz funktsiyani lotin ta'rifiga ulashimiz kerak.
\[Z'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{Z\left( {t + h} \right) - Z\left( t \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {3\left( {t + h} \right) - 4} - \sqrt {3t - 4} }}{h}\]Birinchi funktsiyani baholashni to'g'ri baholaganingizga ishonch hosil qiling. Bu o'quvchilar ushbu muammolarni hal qilishda eng ko'p uchraydigan xatolardan biridir.
2-qadam
Keyinchalik biz suratni ratsionalizatsiya qilishimiz kerak.
\[Z'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {3\left( {t + h} \right) - 4} - \sqrt {3t - 4} } \right)}}{h}\frac{{\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}}{{\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}}\]3-qadam
Endi biz qilishimiz kerak bo'lgan narsa - bu algebra (va bu erda biroz chalkash bo'ladi, lekin bu bunday turdagi muammolar uchun odatiy hol) va biz tugatamiz.
\[\begin{align*}Z'\left( t \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{3\left( {t + h} \right) - 4 - \left( {3t - 4} \right)}}{{h\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{3t + 3h - 4 - 3t + 4}}{{h\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{3h}}{{h\left( {\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{3}{{\sqrt {3\left( {t + h} \right) - 4} + \sqrt {3t - 4} }} = \frac{3}{{2\sqrt {3t - 4} }}\end{align*}\]Bu erda suratni ko'paytirishda ehtiyot bo'ling. Ikkinchi atamada minus belgisini (yoki qavsni) yo'qotish oson. Bu talabalarning juda keng tarqalgan xatosi.
Bu funktsiyaning hosilasi u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{Z'\left( t \right) = \frac{3}{{2\sqrt {3t - 4} }}}}\]