\(f\left( x \right)\) \(x = a\) da farqlanishi mumkinligi sababli biz buni bilamiz

\[f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\]

mavjud. Bu bizga birozdan keyin kerak bo'ladi.

Agar biz \(x \ne a\) deb faraz qilsak, quyidagilarni yozishimiz mumkin:

\[f\left( x \right) - f\left( a \right) = \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {x - a} \right)\]

Keyin chegaralarning asosiy xususiyatlari bizda borligini aytadi:

\[\begin{align*}\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) & = \mathop {\lim }\limits_{x \to a} \left[ {\frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {x - a} \right)} \right]\\ & = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\,\,\,\mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\end{align*}\]

O'ngdagi birinchi chegara yuqorida aytib o'tganimizdek, faqat \(f'\left( a \right)\) va ikkinchi chegara aniq nolga teng va shuning uchun,

\[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( a \right) \cdot 0 = 0\]

OK, biz \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\) ekanligini isbotlashga muvaffaq bo'ldik. Lekin bu qanday qilib \(f\left( x \right)\) \(x = a\) da uzluksiz ekanligini isbotlashga yordam beradi?

Quyidagilardan boshlaylik.

\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + f\left( a \right) - f\left( a \right)} \right]\]

E'tibor bering, biz o'ng tomonda nolga qo'shdik. Bir oz qayta yozish va chegara xususiyatlaridan foydalanish,

\[\begin{align*}\mathop {\lim }\limits_{x \to a} f\left( x \right) & = \mathop {\lim }\limits_{x \to a} \left[ {f\left( a \right) + f\left( x \right) - f\left( a \right)} \right]\\ & = \mathop {\lim }\limits_{x \to a} f\left( a \right) + \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - f\left( a \right)} \right]\end{align*}\]

Endi biz yuqorida \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\) ekanligini isbotladik va \(f\left( a \right)\) doimiy bo'lgani uchun biz \(\mathop {\lim }\limits_{x \to a} f\left( a \right) = f\left( a \right)\) ekanligini ham bilamiz va bu shunday bo'ladi,

\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = \mathop {\lim }\limits_{x \to a} f\left( a \right) + 0 = f\left( a \right)\]

Yoki, boshqacha qilib aytganda, \[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\], lekin \(f\left( x \right)\) \(x = a\) da uzluksiz bo'lishi uchun aynan mana shu ma'noni anglatadi va biz tugatdik.

Bu hosila ta'rifi yordamida isbotlash uchun etarlicha oson. Biz ikkita funktsiyaning yig'indisidan boshlaymiz. Avval yig'indini hosila ta'rifiga kiriting va hisoblagichni biroz qayta yozing.

\[\begin{align*}{\left( {f\left( x \right) + g\left( x \right)} \right)^\prime } & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) + g\left( {x + h} \right) - \left( {f\left( x \right) + g\left( x \right)} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right) + g\left( {x + h} \right) - g\left( x \right)}}{h}\end{align*}\]

Endi kasrni ikkiga bo'ling va yig'indining chegarasi chegaralar yig'indisi ekanligini eslang. Ushbu faktdan foydalanib, biz ikkala funktsiyaning har biri uchun hosila ta'rifi bilan yakunlanganimizni ko'ramiz.

\[\begin{align*}{\left( {f\left( x \right) + g\left( x \right)} \right)^\prime } & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} + \mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h}\\ & = f'\left( x \right) + g'\left( x \right)\end{align*}\]

Ikki funktsiyaning farqining isboti deyarli bir xil, shuning uchun biz buni hech qanday izohsiz keltiramiz.

\[\begin{align*}{\left( {f\left( x \right) - g\left( x \right)} \right)^\prime } & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - g\left( {x + h} \right) - \left( {f\left( x \right) - g\left( x \right)} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right) - \left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} - \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h}\\ & = f'\left( x \right) - g'\left( x \right)\end{align*}\]

Bu xususiyatni ta'rifdan foydalanib isbotlash juda oson, agar siz doimiyni chegaradan chiqarib yuborishimiz mumkinligini eslasangiz. Mana bu mulk uchun ish.

\[{\left( {cf\left( x \right)} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{cf\left( {x + h} \right) - cf\left( x \right)}}{h} = c\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = cf'\left( x \right)\]

Buni hosila ta'rifi yordamida isbotlash juda oson, shuning uchun \(f\left( x \right) = c\) ni aniqlang va hosila ta'rifidan foydalaning.

\[f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{c - c}}{h} = \mathop {\lim }\limits_{h \to 0} 0 = 0\]

Yuqorida ta'kidlab o'tilganidek, bu holda \(n\) musbat butun son deb taxmin qilishimiz kerak. Ushbu teoremada hosila va binom teoremasining ta'rifidan foydalanamiz. Binom teoremasi bizga shuni aytadi:

\[\begin{align*}{\left( {a + b} \right)^n} & = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right){a^{n - k}}{b^k}} \\ & = {a^n} + \left( {\begin{array}{*{20}{c}}n\\1\end{array}} \right){a^{n - 1}}b + \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right){a^{n - 2}}{b^2} + \left( {\begin{array}{*{20}{c}}n\\3\end{array}} \right){a^{n - 3}}{b^3} + \cdots + \left( {\begin{array}{*{20}{c}}n\\{n - 1}\end{array}} \right)a{b^{n - 1}} + \left( {\begin{array}{*{20}{c}}n\\n\end{array}} \right){b^n}\\ & = {a^n} + n{a^{n - 1}}b + \frac{{n\left( {n - 1} \right)}}{{2!}}{a^{n - 2}}{b^2} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{a^{n - 3}}{b^3} + \cdots + na{b^{n - 1}} + {b^n}\end{align*}\]

qayerda,

\[\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right) = \frac{{n!}}{{k!\left( {n - k} \right)!}}\]

binomial koeffitsientlar deyiladi va \(n! = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( 2 \right)\left( 1 \right)\) faktorialdir.

Shunday qilib, keling, ushbu dalilning tafsilotlarini ko'rib chiqaylik. Birinchidan, \(f\left( x \right) = {x^n}\) ni hosila ta'rifiga ulang va birinchi atamani kengaytirish uchun Binom teoremasidan foydalaning.

\[\begin{align*}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^n} - {x^n}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^n} + n{x^{n - 1}}h + \frac{{n\left( {n - 1} \right)}}{{2!}}{x^{n - 2}}{h^2} + \cdots + nx{h^{n - 1}} + {h^n}} \right) - {x^n}}}{h}\end{align*}\]

Endi e'tibor bering, biz \({x^n}\) ni bekor qilishimiz mumkin va keyin suratdagi har bir atamada \(h\) bo'ladi, ular koeffitsientlarga ajratilishi va keyin maxrajdagi \(h\) ga qarshi bekor qilinishi mumkin. Shu nuqtada biz chegarani baholashimiz mumkin.

\[\begin{align*}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{n{x^{n - 1}}h + \frac{{n\left( {n - 1} \right)}}{{2!}}{x^{n - 2}}{h^2} + \cdots + nx{h^{n - 1}} + {h^n}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} n{x^{n - 1}} + \frac{{n\left( {n - 1} \right)}}{{2!}}{x^{n - 2}}h + \cdots + nx{h^{n - 2}} + {h^{n - 1}}\\ & = n{x^{n - 1}}\end{align*}\]

Ushbu dalil uchun biz yana \(n\) ni musbat butun son bo'lishini cheklashimiz kerak. Bunday holda, agar biz \(f\left( x \right) = {x^n}\) ni aniqlasak, hosila ta'rifining muqobil chegara shaklidan bilamizki, \(f'\left( a \right)\) hosilasi quyidagicha beriladi:

\[f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}} = \mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}}\]

Endi bizda quyidagi formula bor,

\[{x^n} - {a^n} = \left( {x - a} \right)\left( {{x^{n - 1}} + a{x^{n - 2}} + {a^2}{x^{n - 3}} + \cdots + {a^{n - 3}}{x^2} + {a^{n - 2}}x + {a^{n - 1}}} \right)\]

Agar xohlasangiz, ikkita omilni shunchaki ko'paytirish orqali buni tasdiqlashingiz mumkin. Shuningdek, ikkinchi omilda jami \(n\) atamalari mavjudligiga e'tibor bering (bu biroz muhim bo'ladi).

Agar biz buni lotin formulasiga kiritsak, biz \(x - a\) ni bekor qilishimiz va chegarani hisoblashimiz mumkinligini ko'ramiz.

\[\begin{align*}f'\left( a \right) & = \mathop {\lim }\limits_{x \to a} \frac{{\left( {x - a} \right)\left( {{x^{n - 1}} + a{x^{n - 2}} + {a^2}{x^{n - 3}} + \cdots + {a^{n - 3}}{x^2} + {a^{n - 2}}x + {a^{n - 1}}} \right)}}{{x - a}}\\ & = \mathop {\lim }\limits_{x \to a} {x^{n - 1}} + a{x^{n - 2}} + {a^2}{x^{n - 3}} + \cdots + {a^{n - 3}}{x^2} + {a^{n - 2}}x + {a^{n - 1}}\\ & = {a^{n - 1}} + a{a^{n - 2}} + {a^2}{a^{n - 3}} + \cdots + {a^{n - 3}}{a^2} + {a^{n - 2}}a + {a^{n - 1}}\\ & = n{a^{n - 1}}\end{align*}\]

Har bir atamadagi ko'rsatkichlarni birlashtirgandan so'ng, biz bir xil atamani olishimizni ko'rishimiz mumkin. Shunday qilib, ikkinchi omilda \(n\) atamalari mavjudligini eslab, biz da'vo qilgan narsaga erishganimizni ko'rishimiz mumkin.

Buni to'liq tugatish uchun biz shunchaki \(a\) ni \(x\) bilan almashtiramiz:

\[f'\left( x \right) = n{x^{n - 1}}\]

Ushbu dalilda biz endi \(n\) ni musbat butun son bo'lishini cheklashimiz shart emas. Endi u har qanday haqiqiy raqam bo'lishi mumkin. Biroq, bu dalil, shuningdek, siz Hosil bo'limni to'liq o'qiganingizni taxmin qiladi. Xususan, ikkalasi ham kerak Yashirin farqlash va Logarifmik farqlash. Agar siz ushbu bo'limlarni o'qimagan bo'lsangiz va tushunmagan bo'lsangiz, unda bu dalil siz uchun hech qanday ma'noga ega bo'lmaydi.

Shunday qilib, logarifmik farqlashni sozlash uchun avval \(y = {x^n}\) ni aniqlaymiz, keyin ikkala tomonning jurnalini olamiz, logarifm xususiyatlaridan foydalangan holda o'ng tomonni soddalashtiramiz va so'ngra yashirin differentsiatsiya yordamida farqlaymiz.

\[\begin{align*}\ln y & = \ln {x^n}\\ \ln y & = n\ln x\\ \frac{{y'}}{y} & = n\frac{1}{x}\end{align*}\]

Va nihoyat, biz qilishimiz kerak bo'lgan yagona narsa \(y'\) ni hal qilish va keyin \(y\) ni almashtirishdir.

\[y' = y\frac{n}{x} = {x^n}\left( {\frac{n}{x}} \right) = n{x^{n - 1}}\]

Bu dalil birinchi marta ko'rganingizda biroz qiyin bo'lishi mumkin, shuning uchun bu erda biroz ehtiyot bo'laylik. Biz birinchi navbatda hosila ta'rifidan foydalanamiz.

\[{\left( {f\,g} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right)g\left( {x + h} \right) - f\left( x \right)g\left( x \right)}}{h}\]

Tashqi tomondan, bu biz uchun hech narsa qilmaydi. Dalilni olish uchun avvalo narsalarni biroz manipulyatsiya qilishimiz kerak. Biz nima qilamiz, ayirish va hisoblagichga \(f\left( {x + h} \right)g\left( x \right)\) qo'shish. E'tibor bering, biz bu erda nol qo'shmoqdamiz, chunki bu ikki shart bekor qilinadi. Bu bizga beradi,

\[{\left( {f\,g} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right)g\left( {x + h} \right) - f\left( {x + h} \right)g\left( x \right) + f\left( {x + h} \right)g\left( x \right) - f\left( x \right)g\left( x \right)}}{h}\]

E'tibor bering, biz suratning o'rtasiga ikkita atama qo'shdik. Yozilganidek, biz chegarani ikki qismga bo'lishimiz mumkin. Birinchi qismdan biz \(f\left( {x + h} \right)\) ni, ikkinchi qismdan esa \(g\left( x \right)\) ni faktor qilishimiz mumkin. Buni qilish,

\[\begin{align*}{\left( {f\,g} \right)^\prime } & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right)\left( {g\left( {x + h} \right) - g\left( x \right)} \right)}}{h} + \mathop {\lim }\limits_{h \to 0} \frac{{g\left( x \right)\left( {f\left( {x + h} \right) - f\left( x \right)} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} f\left( {x + h} \right)\frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} + \mathop {\lim }\limits_{h \to 0} g\left( x \right)\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\end{align*}\]

Ushbu nuqtada yozish uchun chegara xususiyatlaridan foydalanishimiz mumkin,

\[{\left( {f\,g} \right)^\prime } = \left( {\mathop {\lim }\limits_{h \to 0} f\left( {x + h} \right)} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h}} \right) + \left( {\mathop {\lim }\limits_{h \to 0} g\left( x \right)} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}} \right)\]

Bu erda individual chegaralar,

\[\begin{align*}\mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} & = g'\left( x \right) & \hspace{0.75in} & \mathop {\lim }\limits_{h \to 0} g\left( x \right) = g\left( x \right)\\ \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} & = f'\left( x \right) & \hspace{0.75in} & \mathop {\lim }\limits_{h \to 0} f\left( {x + h} \right) = f\left( x \right)\end{align*}\]

Chapdagi ikkita chegara mos ravishda \(g\left( x \right)\) va \(f\left( x \right)\) uchun hosila ta'rifidan boshqa narsa emas. O'ngdagi yuqori chegara biroz qiyin ko'rinadi, lekin doimiy chegara chegarasi faqat doimiy ekanligini unutmang. Bunday holda, chegara faqat \(h\) ning nolga o'tishiga ruxsat berish bilan bog'liq. Bu erda asosiy narsa \(h\) ni o'zgartirish \(x\) ni o'zgartirmasligini tan olishdir va bu chegaraga kelsak, \(g\left( x \right)\) doimiy hisoblanadi. E'tibor bering, funktsiya, ehtimol, doimiy emas, lekin chegaraga kelsak funktsiyani doimiy deb hisoblash mumkin. Biz \(h = 0\)-ni funktsiyaga ulash orqali o'ngdagi pastki chegarani olamiz

Bularning barchasini oxirgi bosqichga ulash bizga beradi,

\[{\left( {f\,g} \right)^\prime } = f\left( x \right)g'\left( x \right) + g\left( x \right)f'\left( x \right)\]

Bu tezroq dalil, lekin siz o'qiganingizni va tushunganingizni taxmin qiladi Yashirin farqlash va Logarifmik farqlash bo'limlar. Agar buni qilmagan bo'lsangiz, bu dalil siz uchun unchalik ma'noli bo'lmaydi.

Avval \(y\) mahsulotiga qo'ng'iroq qiling va ikkala tomonning jurnalini oling va o'ng tomonda logarifmlar xususiyatidan foydalaning.

\[\begin{align*}y & = f\left( x \right)g\left( x \right)\\ \ln \left( y \right) & = \ln \left( {f\left( x \right)g\left( x \right)} \right) = \ln f\left( x \right) + \ln g\left( x \right)\end{align*}\]

Keyin ikkala tomonning hosilasini olamiz va \(y'\) ni hal qilamiz.

\[\frac{{y'}}{y} = \frac{{f'\left( x \right)}}{{f\left( x \right)}} + \frac{{g'\left( x \right)}}{{g\left( x \right)}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y' = y\left( {\frac{{f'\left( x \right)}}{{f\left( x \right)}} + \frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right)\]

Va nihoyat, biz qilishimiz kerak bo'lgan yagona narsa \(y\)-ga ulanish va keyin buni qavs orqali ko'paytirish va biz mahsulot qoidasini olamiz.

\[y' = f\left( x \right)g\left( x \right)\left( {\frac{{f'\left( x \right)}}{{f\left( x \right)}} + \frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\left( {fg} \right)^\prime } = g\left( x \right)f'\left( x \right) + f\left( x \right)g'\left( x \right)\]

Avval qismni hosila ta'rifiga kiriting va qismni biroz qayta yozing.

\[\begin{align*}{\left( {\frac{f}{g}} \right)^\prime } & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{f\left( {x + h} \right)}}{{g\left( {x + h} \right)}} - \frac{{f\left( x \right)}}{{g\left( x \right)}}}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\,\,\frac{{f\left( {x + h} \right)g\left( x \right) - f\left( x \right)g\left( {x + h} \right)}}{{g\left( {x + h} \right)g\left( x \right)}}\end{align*}\]

Hayotimizni biroz osonlashtirish uchun biz \(h\) ni \(\frac{1}{h}\) sifatida birinchi qadam maxrajiga o'tkazdik. Numeratorni ham bitta ratsional ifoda sifatida yozdik. Ushbu dalil ishlashi uchun bu qadam talab qilinadi.

Endi, keyingi qadam uchun ayirish va hisoblagichga \(f\left( x \right)g\left( x \right)\) qo'shish kerak bo'ladi.

\[{\left( {\frac{f}{g}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\,\,\frac{{f\left( {x + h} \right)g\left( x \right) - f\left( x \right)g\left( x \right) + f\left( x \right)g\left( x \right) - f\left( x \right)g\left( {x + h} \right)}}{{g\left( {x + h} \right)g\left( x \right)}}\]

Keyingi qadam narsalarni biroz qayta yozishdir,

\[{\left( {\frac{f}{g}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\,\,\frac{{f\left( {x + h} \right)g\left( x \right) - f\left( x \right)g\left( x \right) + f\left( x \right)g\left( x \right) - f\left( x \right)g\left( {x + h} \right)}}{h}\]

E'tibor bering, biz qilgan yagona narsa ikkita maxrajni almashtirish edi. Biz kasrlarni ko'paytirayotganimiz uchun buni qila olamiz.

Keyinchalik, katta kasrni quyidagicha ajratish mumkin.

\[{\left( {\frac{f}{g}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\,\,\left( {\frac{{f\left( {x + h} \right)g\left( x \right) - f\left( x \right)g\left( x \right)}}{h} + \frac{{f\left( x \right)g\left( x \right) - f\left( x \right)g\left( {x + h} \right)}}{h}} \right)\]

Birinchi kasrda biz \(g\left( x \right)\) ni, ikkinchisida esa \( - f\left( x \right)\) ni ajratamiz. Bu beradi,

\[{\left( {\frac{f}{g}} \right)^\prime } = \mathop {\lim }\limits_{h \to 0} \frac{1}{{g\left( {x + h} \right)g\left( x \right)}}\,\,\left( {g\left( x \right)\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} - f\left( x \right)\frac{{g\left( {x + h} \right) - g\left( x \right)}}{h}} \right)\]

Endi biz limitlarning asosiy xususiyatlaridan foydalanib, buni quyidagicha yozishimiz mumkin:

\[\begin{align*}{\left( {\frac{f}{g}} \right)^\prime } & = \frac{1}{{\mathop {\lim }\limits_{h \to 0} g\left( {x + h} \right)\mathop {\lim }\limits_{h \to 0} g\left( x \right)}}\,\left( {\left( {\mathop {\lim }\limits_{h \to 0} g\left( x \right)} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}} \right) - } \right.\\ & \hspace{2.25in}\left. {\left( {\mathop {\lim }\limits_{h \to 0} f\left( x \right)} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h}} \right)} \right)\end{align*}\]

Shaxsiy chegaralar,

\[\begin{align*}\mathop {\lim }\limits_{h \to 0} \frac{{g\left( {x + h} \right) - g\left( x \right)}}{h} & = g'\left( x \right) & \hspace{0.5in} \mathop {\lim }\limits_{h \to 0} g\left( {x + h} \right) & = g\left( x \right) & \hspace{0.5in} & \mathop {\lim }\limits_{h \to 0} g\left( x \right) = g\left( x \right)\\ \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} & = f'\left( x \right) & \hspace{0.5in} \mathop {\lim }\limits_{h \to 0} f\left( x \right) & = f\left( x \right) & & \end{align*}\]

Har bir qatordagi dastlabki ikkita chegara mos ravishda \(g\left( x \right)\) va \(f\left( x \right)\) uchun hosila ta'rifidan boshqa narsa emas. Yuqori qatordagi o'rta chegarani biz \(h = 0\) ni ulash orqali olamiz. Har bir qatordagi oxirgi chegara biroz qiyin bo'lib tuyulishi mumkin. Eslatib o'tamiz, doimiy chegara faqat doimiydir. Cheklov faqat \(h\) ning tegishli \(g\left( x \right)\) va \(f\left( x \right)\) are konstantalari bo'yicha nolga o'tishiga ruxsat berish bilan bog'liq, chunki \(h\) o'zgarishi o'zgarmaydi. \(x\). E'tibor bering, funktsiya, ehtimol, doimiy emas, lekin chegaraga kelsak, funktsiyani doimiy deb hisoblash mumkin.

Cheklovlarni kiritish va ba'zi qayta tartibga solish,

\[\begin{align*}{\left( {\frac{f}{g}} \right)^\prime } & = \frac{1}{{g\left( x \right)g\left( x \right)}}\,\,\left( {g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)} \right)\\ & = \frac{{f'\,g - f\,g'}}{{{g^2}}}\end{align*}\]

Ko'rsatkich qoidasi mavjud.

Endi logarifmik differentsiatsiya yordamida isbot qilaylik. Biz birinchi navbatda \(y\) koeffitsientini chaqiramiz, ikkala tomonning jurnalini olamiz va o'ng tomondagi jurnallar xususiyatidan foydalanamiz.

\[\begin{align*}y & = \frac{{f\left( x \right)}}{{g\left( x \right)}}\\ \ln y & = \ln \left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \ln f\left( x \right) - \ln g\left( x \right)\end{align*}\]

Keyin ikkala tomonning hosilasini olamiz va \(y'\) ni hal qilamiz.

\[\frac{{y'}}{y} = \frac{{f'\left( x \right)}}{{f\left( x \right)}} - \frac{{g'\left( x \right)}}{{g\left( x \right)}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}y' = y\left( {\frac{{f'\left( x \right)}}{{f\left( x \right)}} - \frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right)\]

Keyin \(y\) ni ulang va qism qoidasini olish uchun biroz soddalashtiring.

\[\begin{align*}y' & = \frac{{f\left( x \right)}}{{g\left( x \right)}}\left( {\frac{{f'\left( x \right)}}{{f\left( x \right)}} - \frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right)\\ & = \frac{{f'\left( x \right)}}{{g\left( x \right)}} - \frac{{g'\left( x \right)f\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\\ & = \frac{{f'\left( x \right)g\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} - \frac{{f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} = \frac{{f'\left( x \right)g\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{align*}\]

Biz dalilni \(u = g\left( x \right)\) ga ta'rif berishdan boshlaymiz va ushbu ta'rif nuqtai nazaridan bizdan isbotlashimiz so'ralgan narsa ekanligini ko'ramiz,

\[\frac{d}{{dx}}\left[ {f\left( u \right)} \right] = f'\left( u \right)\frac{{du}}{{dx}}\]

Keling, \(u\left( x \right)\) hosilasini ko'rib chiqaylik (yana shuni yodda tutingki, biz \(u = g\left( x \right)\) ni aniqlaganmiz va shuning uchun \(u\) haqiqatan ham \(x\) funktsiyasidir) biz borligini bilamiz, chunki biz \(g\left( x \right)\) ni differentsial deb hisoblaymiz. Ta'rifga ko'ra, bizda

\[u'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{u\left( {x + h} \right) - u\left( x \right)}}{h}\]

Shuni ham yodda tutingki,

\[\mathop {\lim }\limits_{h \to 0} \left( {\frac{{u\left( {x + h} \right) - u\left( x \right)}}{h} - u'\left( x \right)} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{u\left( {x + h} \right) - u\left( x \right)}}{h} - \mathop {\lim }\limits_{h \to 0} u'\left( x \right) = u'\left( x \right) - u'\left( x \right) = 0\]

Endi aniqlang,

\[v\left( h \right) = \left\{ {\begin{array}{*{20}{l}}{\displaystyle \frac{{u\left( {x + h} \right) - u\left( x \right)}}{h} - u'\left( x \right)}&{{\mbox{ if }}h \ne 0}\\0&{{\mbox{ if }}h = 0}\end{array}} \right.\]

va \(\mathop {\lim }\limits_{h \to 0} v\left( h \right) = 0 = v\left( 0 \right)\) va shunga o'xshash \(v\left( h \right)\) \(h = 0\) da uzluksiz ekanligiga e'tibor bering.

Endi biz \(h \ne 0\) deb faraz qilsak, \(v\left( h \right)\) ta'rifini qayta yozishimiz mumkin:

\[\begin{equation}u\left( {x + h} \right) = u\left( x \right) + h\left( {v\left( h \right) + u'\left( x \right)} \right)\label{eq:eq1}\end{equation}\]

Endi e'tibor bering, \(\eqref{eq:eq1}\) haqiqatda, agar biz \(h = 0\) ga ruxsat bergan bo'lsak ham amal qiladi va shuning uchun \(h\) ning istalgan qiymati uchun amal qiladi.

Keyinchalik, biz \(f\left( x \right)\) farqlanishini bilganimiz sababli, shunga o'xshash narsani qilishimiz mumkin. Biroq, biz bu erda birozdan keyin aniq bo'ladigan sabablarga ko'ra boshqa harflar/o'zgaruvchilar to'plamidan foydalanamiz. Shunday qilib, aniqlang,

\[w\left( k \right) = \left\{ {\begin{array}{*{20}{l}}{\displaystyle \frac{{f\left( {z + k} \right) - f\left( z \right)}}{k} - f'\left( z \right)}&{{\mbox{ if }}k \ne 0}\\0&{{\mbox{ if }}k = 0}\end{array}} \right.\]

biz yuqorida qilgan shunga o'xshash argumentdan o'tishimiz mumkin, shuning uchun \(w\left( k \right)\) \(k = 0\) da uzluksiz ekanligini ko'rsating va bu,

\[\begin{equation}f\left( {z + k} \right) = f\left( z \right) + k\left( {w\left( k \right) + f'\left( z \right)} \right) \label{eq:eq2} \end{equation}\]

Bu yerda turli harflar haqida hayajonlanmang, biz faqat \(h\) o‘rniga \(k\) dan foydalandik va \(x = z\) ga ruxsat berdik. Bu erda hech qanday ajoyib narsa yo'q, lekin harflarni almashtirish yo'lda foydali bo'ladi.

Xo'sh, shu paytgacha biz zanjir qoidasini isbotlashga yaqinlashtiradigan hech narsa qilmaganga o'xshaymiz. Yuqoridagi ish bizning isbotimizda juda muhim bo'lib chiqadi, ammo keling, dalilga o'tamiz.

Bu erda qilishimiz kerak bo'lgan narsa hosila ta'rifidan foydalanish va quyidagi chegarani baholashdir.

\[\begin{equation}\frac{d}{{dx}}\left[ {f\left[ {u\left( x \right)} \right]} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{f\left[ {u\left( {x + h} \right)} \right] - f\left[ {u\left( x \right)} \right]}}{h} \label{eq:eq3} \end{equation}\]

E'tibor bering, agar biz \(u\) o'rniga \(u\left( x \right)\) dan foydalansak ham, yozuv biroz chalkash bo'lsa ham, bu erda \(u\) haqiqatan ham \(x\) funktsiyasi ekanligini eslatib o'tishimiz kerak.

Keling, \(u\left( {x + h} \right)\)-ni qayta yozish uchun \(\eqref{eq:eq1}\)-dan foydalanamiz va ha, yozuv yoqimsiz bo'ladi, lekin biz buni hal qilishimiz kerak. \(\eqref{eq:eq1}\) dan foydalangan holda, yuqoridagi chegaradagi hisoblagich quyidagicha bo'ladi:

\[f\left[ {u\left( {x + h} \right)} \right] - f\left[ {u\left( x \right)} \right] = f\left[ {u\left( x \right) + h\left( {v\left( h \right) + u'\left( x \right)} \right)} \right] - f\left[ {u\left( x \right)} \right]\]

Agar biz \(z = u\left( x \right)\) va \(k = h\left( {v\left( h \right) + u'\left( x \right)} \right)\) ni aniqlasak, buni quyidagicha yozish uchun \(\eqref{eq:eq2}\) dan foydalanishimiz mumkin:

\[\begin{align*}f\left[ {u\left( {x + h} \right)} \right] - f\left[ {u\left( x \right)} \right] & = f\left[ {u\left( x \right) + h\left( {v\left( h \right) + u'\left( x \right)} \right)} \right] - f\left[ {u\left( x \right)} \right]\\ & = f\left[ {u\left( x \right)} \right] + h\left( {v\left( h \right) + u'\left( x \right)} \right)\left( {w\left( k \right) + f'\left[ {u\left( x \right)} \right]} \right) - f\left[ {u\left( x \right)} \right]\\ & = h\left( {v\left( h \right) + u'\left( x \right)} \right)\left( {w\left( k \right) + f'\left[ {u\left( x \right)} \right]} \right)\end{align*}\]

E'tibor bering, biz ishlarni biroz soddalashtirish uchun \(f\left[ {u\left( x \right)} \right]\)-ni bekor qilishga muvaffaq bo'ldik. Shuni ham yodda tutingki, \(w\left( k \right)\) bu yerda tartibsizlikni minimallashtirish uchun ataylab shunday qoldirilgan, shunchaki esda tutingki, \(k = h\left( {v\left( h \right) + u'\left( x \right)} \right)\) bu yerda birozdan keyin muhim bo‘ladi. Endi orqaga qaytaylik va bularning barchasi bizning chegaramiz \(\eqref{eq:eq3}\) hisoblagichi ekanligini eslaylik. Buni \(\eqref{eq:eq3}\) ga ulash, beradi,

\[\begin{align*}\frac{d}{{dx}}\left[ {f\left[ {u\left( x \right)} \right]} \right] & = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {v\left( h \right) + u'\left( x \right)} \right)\left( {w\left( k \right) + f'\left[ {u\left( x \right)} \right]} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \left( {v\left( h \right) + u'\left( x \right)} \right)\left( {w\left( k \right) + f'\left[ {u\left( x \right)} \right]} \right)\end{align*}\]

\(h\) bekor qilinganiga e'tibor bering. Keyin \(k = h\left( {v\left( h \right) + u'\left( x \right)} \right)\) va shunga o'xshash narsalarni eslang,

\[\mathop {\lim }\limits_{h \to 0} k = \mathop {\lim }\limits_{h \to 0} h\left( {v\left( h \right) + u'\left( x \right)} \right) = 0\]

Ammo, agar \(\mathop {\lim }\limits_{h \to 0} k = 0\), biz baribir \(k\) ni aniqlagan bo'lsak, \(w\) ta'rifi va biz \(w\left( k \right)\) ning \(k = 0\) da uzluksiz ekanligini bilishimiz bilan biz buni ham bilamiz,

\[\mathop {\lim }\limits_{h \to 0} w\left( k \right) = w\left( {\mathop {\lim }\limits_{h \to 0} k} \right) = w\left( 0 \right) = 0\]

Shuningdek, \(\mathop {\lim }\limits_{h \to 0} v\left( h \right) = 0\) ni eslang. Bu faktlarning barchasidan foydalanib, bizning chegaramiz bo'ladi,

\[\begin{align*}\frac{d}{{dx}}\left[ {f\left[ {u\left( x \right)} \right]} \right] & = \mathop {\lim }\limits_{h \to 0} \left( {v\left( h \right) + u'\left( x \right)} \right)\left( {w\left( k \right) + f'\left[ {u\left( x \right)} \right]} \right)\\ & = u'\left( x \right)f'\left[ {u\left( x \right)} \right]\\ & = f'\left[ {u\left( x \right)} \right]\frac{{du}}{{dx}}\end{align*}\]

Aynan shu narsani isbotlashimiz kerak edi va shuning uchun biz tugatdik.