3.12-bo'lim : Yuqori tartibli hosilalar
3. \(\displaystyle f\left( x \right) = 4\,\sqrt[5]{{{x^3}}} - \frac{1}{{8{x^2}}} - \sqrt x \) ning to‘rtinchi hosilasini aniqlang
Yechim
Bu muammo to'rtta hosila olishdan boshqa narsa emas, shuning uchun har bir qadam biz to'rtinchi darajaga yetguncha har bir keyingi hosilani ko'rsatadi. Funktsiyani tezda qayta yozgandan so'ng, farqlashda yordam berish uchun birinchi hosila:
\[f\left( x \right) = 4\,{x^{\frac{3}{5}}} - \frac{1}{8}{x^{ - 2}} - {x^{\frac{1}{2}}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}f'\left( x \right) = \frac{{12}}{5}\,{x^{ - \,\,\frac{2}{5}}} + \frac{1}{4}{x^{ - 3}} - \frac{1}{2}{x^{ - \,\,\frac{1}{2}}}\]2-qadam
Ikkinchi hosila,
\[f''\left( x \right) = - \frac{{24}}{{25}}\,{x^{ - \,\,\frac{7}{5}}} - \frac{3}{4}{x^{ - 4}} + \frac{1}{4}{x^{ - \,\,\frac{3}{2}}}\]3-qadam
Uchinchi hosila,
\[f'''\left( x \right) = \frac{{168}}{{125}}\,{x^{ - \,\,\frac{{12}}{5}}} + 3{x^{ - 5}} - \frac{3}{8}{x^{ - \,\,\frac{5}{2}}}\]4-qadam
Ushbu muammoning to'rtinchi va oxirgi hosilasi:
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{f^{\left( 4 \right)}}\left( x \right) = - \frac{{2016}}{{625}}\,{x^{ - \,\,\frac{{17}}{5}}} - 15{x^{ - 6}} + \frac{{15}}{{16}}{x^{ - \,\,\frac{7}{2}}}}}\]