Question

помогите срочно нужно ​

Answer 1

$\Big(\dfrac{x^4}{4}\Big)'=\dfrac{1}{4}\cdot 4x^3=x^3\\\\\\\Big(\dfrac{4}{x}\Big)'=4\cdot (-1)\cdot x^{-2}=-\dfrac{4}{x^2}\\\\\\\Big(\dfrac{4}{x^4}\Big)'=4\cdot (-4)\cdot x^{-5}=-\dfrac{16}{x^5}\\\\\\\Big(4\sqrt{x} \Big)'=4\cdot \dfrac{1}{2\sqrt{x}}=\dfrac{2}{\sqrt{x}}\\\\\\(4\sqrt[8]{x})'=4\cdot \dfrac{1}{8}\cdot x^{-\frac{7}{8}}=\dfrac{1}{2\sqrt[8]{x^7}}\\\\\\\Big(\dfrac{4}{\sqrt{x}}\Big)'=4\cdot (-\dfrac{1}{2})\cdot x^{-\frac{3}{2}}=-\dfrac{2}{\sqrt{x^3}}$

$\Big(\dfrac{4}{\sqrt[8]{x^3}}\Big)'=4\cdot (-\dfrac{3}{8})\cdot x^{-\frac{11}{8}}=-\dfrac{3}{2\sqrt[8]{x^{11}}}\\\\\\(7x^9-2ctgx-8x+1)'=63x^8+\dfrac{2}{sin^2x}-8\\\\\\(x^{10}\cdot sinx)'=10x^9\cdot sinx+x^{10}\cdot cosx\\\\\\\Big(\dfrac{x^{10}}{cosx}\Big)'=\dfrac{10x^9\cdot cosx+x^{10}\cdot sinx}{cos^2x}$