Задача 68118 найти производные функции, не понимаю...
Условие
Решение
1) показательной:
[m](a^{u(x)})`=a^{u(x)}\cdot lna\cdot u`(x)[/m]⇒ [m]( 10^{\sqrt{2+x}})`=( 10^{\sqrt{2+x}}\cdot ln10\cdot (\sqrt{2+x})`[/m]
2) степенной:
[m]((u(x))^3)`=3u^2\cdot u`(u(x))[/m]⇒ [m](sin^3(u))`=3\cdot (sin^2u)\cdot (sin u)`[/m],
3) тригонометрической
[m](sinu(x))`=(cosu(x))\cdot u`(x)[/m]
[m]u(x)=\frac{x+\sqrt{1+x^2}}{x}[/m]
По правилу производной частного:
[m]u`(x)=\frac{(x+\sqrt{1+x^2})`\cdot x-(x+\sqrt{1+x^2})\cdot (x)`}{x^2}[/m]
Получаем:
[m]y`=(x^6)`\cdot 10^{\sqrt{2+x}}+x^6\cdot (10^{\sqrt{2+x}})`+3sin^2(\frac{x+\sqrt{1+x^2}}{x})\cdot cos(\frac{x+\sqrt{1+x^2}}{x})\cdot (\frac{x+\sqrt{1+x^2}}{x})`[/m]
[m]y`=(6x^5)\cdot 10^{\sqrt{2+x}}+x^6\cdot (10^{\sqrt{2+x}})\cdot ln10 \cdot (\sqrt{2+x})`+3sin^2(\frac{x+\sqrt{1+x^2}}{x})\cdot cos(\frac{x+\sqrt{1+x^2}}{x})\cdot \frac{(x+\sqrt{1+x^2})`\cdot x-(x+\sqrt{1+x^2})\cdot (x)`}{x^2}[/m]
[m]y`=(6x^5)\cdot 10^{\sqrt{2+x}}+x^6\cdot (10^{\sqrt{2+x}})\cdot\cdot ln10 \frac{1}{2\sqrt{2+x}}\cdot (2+x)`+3sin^2(\frac{x+\sqrt{1+x^2}}{x})\cdot cos(\frac{x+\sqrt{1+x^2}}{x})\cdot \frac{(1+\frac{1}{2\sqrt{1+x^2}}\cdot (1+x^2)`)\cdot x-(x+\sqrt{1+x^2})\cdot 1}{x^2}[/m]
[m]y`=(6x^5)\cdot 10^{\sqrt{2+x}}+x^6\cdot (10^{\sqrt{2+x}})\cdot ln10\cdot \frac{1}{2\sqrt{2+x}}+3sin^2(\frac{x+\sqrt{1+x^2}}{x})\cdot cos(\frac{x+\sqrt{1+x^2}}{x})\cdot \frac{x+\frac{x^2}{\sqrt{1+x^2}}-x-\sqrt{1+x^2})}{x^2}[/m]
[m]y`=(6x^5)\cdot 10^{\sqrt{2+x}}+x^6\cdot (10^{\sqrt{2+x}})\cdot ln10\cdot \frac{1}{2\sqrt{2+x}}+3sin^2(\frac{x+\sqrt{1+x^2}}{x})\cdot cos(\frac{x+\sqrt{1+x^2}}{x})\cdot \frac{x^2-(1+x^2)}{x^2\sqrt{1+x^2}}[/m]
[m]y`=(6x^5)\cdot 10^{\sqrt{2+x}}+x^6\cdot (10^{\sqrt{2+x}})\cdot ln10\cdot \frac{1}{2\sqrt{2+x}}-3sin^2(\frac{x+\sqrt{1+x^2}}{x})\cdot cos(\frac{x+\sqrt{1+x^2}}{x})\cdot \frac{1}{x^2\sqrt{1+x^2}}[/m]