# Thí dụ toán đạo hàm

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

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

• ${\displaystyle f(x)=\sin(x)}$
${\displaystyle f'(x)=\lim _{h\to 0}{\frac {\sin(x+h)-\sin(x)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\sin(h)+\cos(h)\sin(x)-\sin(x)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\sin(h)+(\cos(h)-1)\sin(x)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\sin(h)}{h}}+\lim _{h\to 0}{\frac {(\cos(h)-1)\sin(x)}{h}}}$
${\displaystyle =\cos(x)\times 1+\sin(x)\times 0}$
${\displaystyle =\cos(x)}$

• ${\displaystyle f(x)=\cos(x)}$
${\displaystyle f'(x)=\lim _{h\to 0}{\frac {\cos(x+h)-\cos(x)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)\cos(h)-\sin(h)\sin(x)-\cos(x)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h}}}$
${\displaystyle =\lim _{h\to 0}{\frac {\cos(x)(\cos(h)-1)}{h}}-\lim _{h\to 0}{\frac {\sin(x)\sin(h)}{h}}}$
${\displaystyle =\cos(x)\times 0-\sin(x)\times 1}$
${\displaystyle =-\sin(x)}$