Thí dụ toán đạo hàm

• $f(x)=x^2$

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

$f(x)=x^2$

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

• $f(x)=e^x$

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

$f(x)=e^x$

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

• $f(x)=\sin (x)$

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\sin (x+h)-\sin (x)}{h}$

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

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

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

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

$=\cos (x)$

• $f(x)=\cos (x)$

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos (x)}{h}$

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

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

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

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

$=-\sin (x)$