3.10-bo'lim : Yashirin farqlash
8. \(y'\) ni \(\cos \left( {{x^2} + 2y} \right) + x\,{{\bf{e}}^{{y^{\,2}}}} = 1\) 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. Bu chap tomondagi ikkinchi atama haqiqatan ham \(x\) funksiyalarining mahsuloti ekanligini anglatadi va shuning uchun biz ushbu atamani farqlashda mahsulot qoidasidan foydalanishimiz kerak bo'ladi.
\(x\) ga nisbatan farqlash shuni beradi,
\[ - \left( {2x + 2y'} \right)\sin \left( {{x^2} + 2y} \right) + \,{{\bf{e}}^{{y^{\,2}}}} + 2y\,y'x\,{{\bf{e}}^{{y^{\,2}}}} = 0\]2-qadam
Nihoyat, biz qilishimiz kerak bo'lgan yagona narsa buni \(y'\) uchun hal qilishdir (ba'zi potentsial chalkash algebra bilan).
\[\begin{align*} - 2x\sin \left( {{x^2} + 2y} \right) - 2y'\sin \left( {{x^2} + 2y} \right) + \,{{\bf{e}}^{{y^{\,2}}}} + 2y\,y'x\,{{\bf{e}}^{{y^{\,2}}}} & = 0\\ \left( {2yx\,{{\bf{e}}^{{y^{\,2}}}} - 2\sin \left( {{x^2} + 2y} \right)} \right)y' & = 0 + 2x\sin \left( {{x^2} + 2y} \right) - \,{{\bf{e}}^{{y^{\,2}}}}\\ y' & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2x\sin \left( {{x^2} + 2y} \right) - \,{{\bf{e}}^{{y^{\,2}}}}}}{{2yx\,{{\bf{e}}^{{y^{\,2}}}} - 2\sin \left( {{x^2} + 2y} \right)}}}}\end{align*}\]