Thí dụ toán đạo hàm

• $f(x)=x^{2}$

$f(x+h)=(x+h)^{2}=x^{2}+2xh+h^{2}$

$f(x)=x^{2}$

${\frac {df(x)}{dx}}=f^{'}(x)=\lim _{h\to 0}\sum {\frac {(x+h)^{2}-x^{2}}{h}}=\lim _{h\to 0}2x+h=2x$

• $f(x)=e^{x}$

$f(x+h)=e^{x+h}$

$f(x)=e^{x}$

${\frac {df(x)}{dx}}=f^{'}(x)=\lim _{h\to 0}\sum {\frac {e^{x+h}}{h}}-e^{x}=\lim _{h\to 0}e^{x}(e^{h}-1)=e^{x}\lim _{h\to 0}(e^{h}-1)=e^{x}.1=e^{x}$

• $f(x)=\sin(x)$

$f'(x)=\lim _{h\to 0}{\frac {\sin(x+h)-\sin(x)}{h}}$

$=\lim _{h\to 0}{\frac {\cos(x)\sin(h)+\cos(h)\sin(x)-\sin(x)}{h}}$

$=\lim _{h\to 0}{\frac {\cos(x)\sin(h)+(\cos(h)-1)\sin(x)}{h}}$

$=\lim _{h\to 0}{\frac {\cos(x)\sin(h)}{h}}+\lim _{h\to 0}{\frac {(\cos(h)-1)\sin(x)}{h}}$

$=\cos(x)\times 1+\sin(x)\times 0$

$=\cos(x)$

• $f(x)=\cos(x)$

$f'(x)=\lim _{h\to 0}{\frac {\cos(x+h)-\cos(x)}{h}}$

$=\lim _{h\to 0}{\frac {\cos(x)\cos(h)-\sin(h)\sin(x)-\cos(x)}{h}}$

$=\lim _{h\to 0}{\frac {\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h}}$

$=\lim _{h\to 0}{\frac {\cos(x)(\cos(h)-1)}{h}}-\lim _{h\to 0}{\frac {\sin(x)\sin(h)}{h}}$

$=\cos(x)\times 0-\sin(x)\times 1$

$=-\sin(x)$