Упр.10.4 ГДЗ Мордкович Семенов 11 класс (Алгебра)
а) y=x^3(1+vx); г) y=x^2(vx-2);
б) y=-2x^2(x^3-4); д) y=3x^3(x^2+5);
в) y=4vx·(1-5x^2); е) y=(2+(1/6)x^3)·2vx.
$$y=x^3(1+\sqrt{x})$$
Применим правило производной произведения:
$$y’=(x^3)'(1+\sqrt{x})+x^3(1+\sqrt{x})’$$
$$y’=3x^2(1+\sqrt{x})+x^3\cdot \frac{1}{2\sqrt{x}}$$
$$y’=3x^2+3x^2\sqrt{x}+\frac{x^3}{2\sqrt{x}}=3x^2+3x^2\sqrt{x}+\frac12 x^2\sqrt{x}$$
$$y’=x^2\left(3+\frac72\sqrt{x}\right)$$
$$y=-2x^2(x^3-4)$$
$$y’=-2\bigl((x^2)'(x^3-4)+x^2(x^3-4)’\bigr)$$
$$y’=-2\bigl(2x(x^3-4)+x^2\cdot 3x^2\bigr)$$
$$y’=-4x(x^3-4)-6x^4=-10x^4+16x$$
$$y=4\sqrt{x}(1-5x^2)$$
$$y’=4\left((\sqrt{x})'(1-5x^2)+\sqrt{x}(1-5x^2)’\right)$$
$$y’=4\left(\frac{1}{2\sqrt{x}}(1-5x^2)+\sqrt{x}\cdot(-10x)\right)$$
$$y’=\frac{2(1-5x^2)}{\sqrt{x}}-40x\sqrt{x}$$
$$y’=\frac{2-5x^2-40x^2}{\sqrt{x}}=\frac{2-45x^2}{\sqrt{x}}$$
$$y=x^2(\sqrt{x}-2)$$
$$y’=(x^2)'(\sqrt{x}-2)+x^2(\sqrt{x}-2)’$$
$$y’=2x(\sqrt{x}-2)+x^2\cdot \frac{1}{2\sqrt{x}}$$
$$y’=2x\sqrt{x}-4x+\frac{x^2}{2\sqrt{x}}=2x\sqrt{x}-4x+\frac12 x\sqrt{x}$$
$$y’=x\left(\frac52\sqrt{x}-4\right)$$
$$y=3x^3(x^2+5)$$
$$y’=3\bigl((x^3)'(x^2+5)+x^3(x^2+5)’\bigr)$$
$$y’=3\bigl(3x^2(x^2+5)+x^3\cdot 2x\bigr)$$
$$y’=3(3x^4+15x^2+2x^4)=15x^4+45x^2$$
$$y’=15x^2(x^2+3)$$
$$y=\left(2+\frac16x^3\right)\cdot 2\sqrt{x}$$
$$y’=\left(2+\frac16x^3\right)’2\sqrt{x}+\left(2+\frac16x^3\right)(2\sqrt{x})’$$
$$y’=\frac12x^2\cdot 2\sqrt{x}+\left(2+\frac16x^3\right)\cdot \frac{1}{\sqrt{x}}$$
$$y’=x^2\sqrt{x}+\frac{2+\frac16x^3}{\sqrt{x}}=\frac{x^3}{\sqrt{x}}+\frac{12+x^3}{6\sqrt{x}}$$
$$y’=\frac{7x^3+12}{6\sqrt{x}}$$
Ответ
а) $$y’=x^2\left(3+\frac72\sqrt{x}\right)$$; б) $$y’=-10x^4+16x$$; в) $$y’=\frac{2-45x^2}{\sqrt{x}}$$; г) $$y’=x\left(\frac52\sqrt{x}-4\right)$$; д) $$y’=15x^2(x^2+3)$$; е) $$y’=\frac{7x^3+12}{6\sqrt{x}}$$.
