Question

Помогите решить первое задание и второе

Answer 1

$1a) \ \frac{\Delta f(x)}{\Delta x} =\frac{f(x+\Delta x)-f(x)}{\Delta x} =\frac{3(x+\Delta x)^2-3x^2}{\Delta x} =\frac{3(2+1)^2-3*2^2}{1}=15\\ \\ 1 \delta) \ \lim_{\Delta x \to 0} \frac{\Delta f(x)}{\Delta x}=\lim_{\Delta x \to 0}\frac{3(x+\Delta x)^2-3x^2}{\Delta x} =\lim_{\Delta x \to 0}\frac{3(2+\Delta x)^2-12}{\Delta x} =\\ \\ =\lim_{\Delta x \to 0}\frac{3(4+4\Delta x+\Delta x^2)-12}{\Delta x} =\lim_{\Delta x \to 0}\frac{12+12\Delta x+3\Delta x^2-12}{\Delta x} =$

$\lim_{\Delta x \to 0}\frac{3\Delta x^2+12\Delta x}{\Delta x} =\lim_{\Delta x \to 0}(3\Delta x+12)=3*0+12=12$

$2) \\a) \ f(x)=x^3+\frac{1}{x^3} -4\sqrt{x} +sin\frac{\pi}{3} =x^3+x^{-3} -4\sqrt{x} +sin\frac{\pi}{3} \\ \\ f'(x)=3x^2-3x^{-4}-4*\frac{1}{2\sqrt{x}} +0=3x^2-\frac{3}{x^4} -\frac{2}{\sqrt{x}} \\ \\\\ \delta ) \ g(x)=(x^2-x)\sqrt{x} \\ \\ g'(x)=(x^2-x)'\sqrt{x}+(x^2-x)(\sqrt{x})'=(2x-1)\sqrt{x} +\frac{x^2-x}{2\sqrt{x}}$

$B) \ t(x)=(\frac{x}{2} -x^3)^4 \\ \\ t'(x)=4(\frac{x}{2} -x^3)^3(\frac{1}{2} -3x^2)$