Question

Помогите с производными

Answer 1

$(x^{n})'=n\cdot x^{n-1}\\\\(\frac{x}{4})'=(\frac{1}{4}\cdot x)'=\frac{1}{4}\\\\(\frac{1}{x^{11}})'=(x^{-11})'=-11\cdot x^{-12}=-\frac{11}{x^{12}}\\\\(x^{12})'=12\cdot x^{11}\\\\(\sqrt{x})'=(x^\frac{1}{2})'=\frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\\\\(\frac{1}{\sqrt{x}})'=(x^{-\frac{1}{2}})'=-\frac{1}{2}\cdot x^{-\frac{3}{2}}=-\frac{1}{2\sqrt{x^3}}\\\\(x^{-6})'=-6\cdot x^{-7}=-\frac{6}{x^7}\\\\(\frac{1}{\sqrt[5]{x}})'=(x^{-\frac{1}{5}})'=-\frac{1}{5}\cdot x^{-\frac{6}{5}}=-\frac{1}{5\sqrt[5]{x^6}}$

$(\sqrt[5]{x})'=(x^{\frac{1}{5} })'=\frac{1}{5}\cdot x^{-\frac{4}{5} }=\frac{1}{5\sqrt[5]{x^4}}\\\\(\sqrt3)'=0\; \; ,\; \; t.k.\; \; \sqrt3=const\\\\(3-2x)'=3'-(2x)'=0-2\cdot x'=-2\cdot 1=-2\\\\(\frac{1}{\sqrt[6]{x^2}})'=(x^{\frac{2}{6}})'=(x^{\frac{1}{3}})'=\frac{1}{3}\cdot x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^2}}$