3.10-bo'lim : Yashirin farqlash
6. \(y'\) ni \({{\bf{e}}^x} - \sin \left( y \right) = x\) uchun yashirin farqlash orqali toping.
Yechim
Birinchidan, biz \(x\) ga nisbatan hamma narsaning hosilasini olishimiz kerak va biz \(y\) haqiqatan ham \(y\left( x \right)\) ekanligini esga olishimiz kerak, shuning uchun biz \(y\) ishtirokidagi atamalarning hosilasini olishda zanjir qoidasidan foydalanishimiz kerak.
\(x\) ga nisbatan farqlash shuni beradi,
\[{{\bf{e}}^x} - \cos \left( y \right)y' = 1\]Farqlashdan keyin kosinusdagi \(y'\) ni unutmang. Shunga qaramay, \(y\) haqiqatan ham \(y\left( x \right)\) va shuning uchun \(\sin \left( y \right)\)-ni farqlashda biz haqiqatan ham \(\sin \left[ {y\left( x \right)} \right]\)-ni farqlaymiz va shuning uchun biz zanjir qoidasi yordamida farqlaymiz!
2-qadam
Va nihoyat, biz qilishimiz kerak bo'lgan yagona narsa buni \(y'\) uchun hal qilishdir.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{y' = \frac{{1 - {{\bf{e}}^x}}}{{ - \cos \left( y \right)}} = \left( {{{\bf{e}}^x} - 1} \right)\sec \left( y \right)}}\]