Question

528. При помощи oпределением, найдите прoизвoдную от функции $y=f(x)$
c)$f(x)=x^{2}-4$
d)$f(x)=2x-x^{2}$

Answer 1

Ответ:

$2x; \quad 2-2x;$

Объяснение:

$c) \quad f(x)=x^{2}-4;$

$f'(x)= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x};$

$f(x+\Delta x )-f(x)=(x+\Delta x)^{2}-4-(x^{2}-4)=x^{2}+2x\Delta x+(\Delta x)^{2}-4-x^{2}+4=$

$=2x\Delta x+(\Delta x)^{2};$

$f'(x)= \lim_{\Delta x \to 0} \frac{2x\Delta x+(\Delta x)^{2}}{\Delta x}=\lim_{\Delta x \to 0} (2x+\Delta x)=2x;$

$d) \quad f(x)=2x-x^{2};$

$f'(x)= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x};$

$f(x+\Delta x)-f(x)=2 \cdot (x+\Delta x)-(x+\Delta x)^{2}-(2x-x^{2})=2x+2\Delta x-$

$-(x^{2}+2x\Delta x+(\Delta x)^{2})-2x+x^{2}=2x-2x+2\Delta x-x^{2}+x^{2}-2x\Delta x-(\Delta x)^{2}=$

$=2\Delta x-2x\Delta x-(\Delta x)^{2};$

$f'(x)= \lim_{\Delta x \to 0} \frac{2\Delta x-2x\Delta x-(\Delta x)^{2}}{\Delta x}=\lim_{\Delta x \to 0} (2-2x-\Delta x)=2-2x;$