Упр.40.10 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

Найдём производные по определению.

1. $$y=\sqrt{x}$$

   $$f(x+\Delta x)=\sqrt{x+\Delta x}$$
   $$\Delta y=f(x+\Delta x)-f(x)=\sqrt{x+\Delta x}-\sqrt{x}$$
   $$\frac{\Delta y}{\Delta x}=\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$$
   $$\frac{\Delta y}{\Delta x}=\frac{(\sqrt{x+\Delta x}-\sqrt{x})(\sqrt{x+\Delta x}+\sqrt{x})}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}$$
   $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\frac{1}{2\sqrt{x}}$$

2. $$y=\frac{1}{x^2}$$

   $$f(x+\Delta x)=\frac{1}{(x+\Delta x)^2}$$
   $$\Delta y=\frac{1}{(x+\Delta x)^2}-\frac{1}{x^2}=\frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2}$$
   $$\Delta y=\frac{x^2-x^2-2x\Delta x-(\Delta x)^2}{x^2(x+\Delta x)^2}=-\frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2}$$
   $$\frac{\Delta y}{\Delta x}=-\frac{2x+\Delta x}{x^2(x+\Delta x)^2}$$
   $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=-\frac{2x}{x^2\cdot x^2}=-\frac{2}{x^3}$$

3. $$y=\sqrt{x}+1$$

   $$f(x+\Delta x)=\sqrt{x+\Delta x}+1$$
   $$\Delta y=\left(\sqrt{x+\Delta x}+1\right)-\left(\sqrt{x}+1\right)=\sqrt{x+\Delta x}-\sqrt{x}$$
   $$\frac{\Delta y}{\Delta x}=\frac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}$$
   $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=\frac{1}{2\sqrt{x}}$$

4. $$y=x^3$$

   $$f(x+\Delta x)=(x+\Delta x)^3=x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3$$
   $$\Delta y=f(x+\Delta x)-f(x)=3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3$$
   $$\frac{\Delta y}{\Delta x}=3x^2+3x\Delta x+(\Delta x)^2$$
   $$\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}=3x^2$$

Ответ

а) $$y’=\frac{1}{2\sqrt{x}}$$; б) $$y’=-\frac{2}{x^3}$$; в) $$y’=\frac{1}{2\sqrt{x}}$$; г) $$y’=3x^2$$.