3.1-bo'lim : Hosilaning ta'rifi
11. hosila ta’rifidan foydalanib, hosilasini toping,
\[f\left( x \right) = \sqrt {1 - 9x} \]Yechim
Avval biz funktsiyani lotin ta'rifiga ulashimiz kerak.
\[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 - 9\left( {x + h} \right)} - \sqrt {1 - 9x} }}{h}\]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.
\[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\sqrt {1 - 9\left( {x + h} \right)} - \sqrt {1 - 9x} } \right)}}{h}\frac{{\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}}{{\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}}\]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.
\[\begin{align*}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 9\left( {x + h} \right) - \left( {1 - 9x} \right)}}{{h\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 9x - 9h - 1 + 9x}}{{h\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{ - 9h}}{{h\left( {\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} } \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - 9}}{{\sqrt {1 - 9\left( {x + h} \right)} + \sqrt {1 - 9x} }} = \frac{{ - 9}}{{2\sqrt {1 - 9x} }}\end{align*}\]Bu erda raqamni 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,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = \frac{{ - 9}}{{2\sqrt {1 - 9x} }}}}\]