Question

Вычислите производную функции по определению: а)f(x)= 2+3x б)f(x)=3x²+7 в) f(x)= (4-x)²

Answer 1

$1)\; \; f(x)=2+3x\\\\f'(x)=lim_{\Delta x\to 0}\, \frac{\Delta f}{\Delta x}=lim_{\Delta x\to 0}\frac{2+3(x+\Delta x)-(2+3x)}{\Delta x}=\\\\=lim_{\Delta x\to 0}= \frac{3\Delta x}{\Delta x} =3\; ;\\\\2)\; \; f(x)=3x^2+7\\\\y'=lim_{\Delta x\to 0} \frac{3(x+\Delta x)^2+7-(3x^2+7)}{\Delta x} =lim \; \frac{6x\cdot \Delta x-3(\Delta x)^2}{\Delta x} =\\\\=lim_{\Delta x\to 0} \frac{\Delta x(6x-3\cdot \Delta x)}{\Delta x} =lim_{\Delta x\to 0}\, (6x-3\cdot \underbrace{\Delta x}_{0})=6x$

$3)\; \; y=(4-x)^2\\\\y'=lim_{\Delta x\to 0} \frac{(4-(x+\Delta x))^2-(4-x)^2}{\Delta x} =lim\; \frac{16-8(x+\Delta x)+(x+\Delta x)^2-(16-8x+x^2)}{\Delta x} \\\\=lim\; \frac{16-8x-8\cdot \Delta x+x^2+2x\cdot \Delta x+(\Delta x)^2-16+8x-x^2}{\Delta } =\\\\=lim_{\Delta x\to 0} \frac{-8\cdot \Delta x+2x\cdot \Delta x+(\Delta x)^2}{\Delta x} =lim_{\Delta x\to 0} \frac{\Delta x(-8+2x+\Delta x)}{\Delta x} =\\\\=lim_{\Delta x\to 0} (2x-8+\underbrace {\Delta x}_{0})=2x-8$