f′(x)=limh→0f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}f′(x)=h→0limhf(x+h)−f(x)
求:f(x)=1xf(x)=\frac{1}{x}f(x)=x1
f′(x)=limh→01x+h−1xhf'(x)=\lim_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}f′(x)=h→0limhx+h1−x1
如果此时用000替换hhh,分母为000失败,把分子通分
f′(x)=limh→0x−(x+h)x(x+h)h=limh→0−hhx(x+h)=limh→0−1x(x+h)f'(x)=\lim_{h\to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h}=\lim_{h\to 0}\frac{-h}{hx(x+h)}=\lim_{h\to 0}\frac{-1}{x(x+h)}f′(x)=h→0limhx(x+h)x−(x+h)=h→0limhx(x+h)−h=h→0limx(x+h)−1
接着用000替换hhh,得到f′(x)=−1x2f'(x)=-\frac{1}{x^2}f′(x)=−x21
也就是ddx(1x)=−1x2\frac{d}{dx}(\frac{1}{x})=-\frac{1}{x^2}dxd(x1)=−x21
求:f(x)=xf(x)=\sqrt{x}f(x)=x
f′(x)=limh→0f(x+h)−f(x)h=limh→0x+h−xhf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}f′(x)=h→0limhf(x+h)−f(x)=h→0limhx+h−x
=limh→0x+h−xh×x+h+xx+h+x=limh→0(x+h)−xh(x+h+x)=limh→01x+h+x=12x=\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\lim_{h\to 0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}=\lim_{h\to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}}=h→0limhx+h−x×x+h+xx+h+x=h→0limh(x+h+x)(x+h)−x=h→0limx+h+x1=2x1