Xo'sh, bizdan asosan bu erda yashirin differentsiatsiya qilish so'raladi va biz hosila ishimizni bajarganimizda \(z\) aslida \(z\left( {x,y} \right)\) deb faraz qilayotganimizni eslaymiz.

Keling, avval \(\frac{{\partial z}}{{\partial x}}\) ni olamiz va bu bizdan \(x\)ga nisbatan farqlashni talab qiladi. Eslatib o‘tamiz, \(x\) va \(z\) o‘z ichiga olgan har qanday mahsulot mahsulot qoidasini talab qiladi, chunki biz \(z\) \(x\) funksiyasi deb taxmin qilamiz. Shuningdek, “ichki” funksiyani farqlashda \(\frac{{\partial z}}{{\partial x}}\) ni olish uchun \(z\) ning har qanday hosilasini to‘g‘ri zanjir qoidasini eslang.

\(x\) ga nisbatan tenglamani farqlash,

\[2x\sin \left( {{y^3}} \right) + {{\bf{e}}^{3z}} + 3\frac{{\partial z}}{{\partial x}}x{{\bf{e}}^{3z}} + 2z\frac{{\partial z}}{{\partial x}}\sin \left( {{z^2}} \right) = - 6\frac{{\partial z}}{{\partial x}}\]

\(\frac{{\partial z}}{{\partial x}}\) ni hal qilish,

\[2x\sin \left( {{y^3}} \right) + {{\bf{e}}^{3z}} = \left( { - 6 - 3x{{\bf{e}}^{3z}} - 2z\sin \left( {{z^2}} \right)} \right)\frac{{\partial z}}{{\partial x}}\hspace{0.5in} \to \hspace{0.5in} \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\partial z}}{{\partial x}} = \frac{{2x\sin \left( {{y^3}} \right) + {{\bf{e}}^{3z}}}}{{ - 6 - 3x{{\bf{e}}^{3z}} - 2z\sin \left( {{z^2}} \right)}}}}\]

2-qadam

Endi biz hamma narsani qaytadan bajarishimiz kerak, bundan tashqari bu safar \(\frac{{\partial z}}{{\partial y}}\) ni topish uchun \(y\)ga nisbatan farqlaymiz. Shunday qilib, farqlash beradi,

\[3{y^2}{x^2}\cos \left( {{y^3}} \right) + 3\frac{{\partial z}}{{\partial y}}x{{\bf{e}}^{3z}} + 2z\frac{{\partial z}}{{\partial y}}sin\left( {{z^2}} \right) = 3 - 6\frac{{\partial z}}{{\partial y}}\]

\(\frac{{\partial z}}{{\partial y}}\) uchun yechish beradi,

\[3{y^2}{x^2}\cos \left( {{y^3}} \right) - 3 = \left( { - 6 - 3x{{\bf{e}}^{3z}} - 2zsin\left( {{z^2}} \right)} \right)\frac{{\partial z}}{{\partial y}}\hspace{0.5in} \to \hspace{0.5in} \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\partial z}}{{\partial y}} = \frac{{3{y^2}{x^2}\cos \left( {{y^3}} \right) - 3}}{{ - 6 - 3x{{\bf{e}}^{3z}} - 2zsin\left( {{z^2}} \right)}}}}\]