Question

Найти производные функций

Answer 1

Ответ:

решение на фотографии

Answer 2

$1)\; \; y=2x^4-\frac{3}{x^3}+8\sqrt[4]{x^3}+5\\\\y'=8x^3-3\cdot (-3)\cdot x^{-4}+8\cdot \frac{3}{4}\cdot x^{-1/4}+0=8x^3+\frac{9}{x^4}+\frac{6}{\sqrt[4]{x}}$

$2)\; \; y=x+e^{-x}\cdot cos4x\\\\y'=1-e^{-x}\cdot cos4x-4\, e^{-x}\cdot sin4x$

$3)\; \; y=\dfrac{2-tg\sqrt{x}}{sin(5x+3)}\\\\y'=\dfrac{-\frac{1}{cos^2\sqrt{x}}\cdot \frac{1}{2\sqrt{x}}\cdot sin(5x+3)-(2-tg\sqrt{x})\cdot 5cos(5x+3)}{sin^2(5x+3)}=\\\\\\=\dfrac{-sin(5x+3)-10\sqrt{x}\cdot cos^2\sqrt{x}\cdot cos(5x+3)\cdot (2-tg\sqrt{x})}{2\sqrt{x}\cdot cos^2\sqrt{x}\cdot sin^2(5x+3)}$

$4)\; \; y=\Big(x^4-ln(2x+1)\Big)^3\\\\y'=3\cdot \Big(x^4-ln(2x+1)\Big)^2\cdot \Big(4x^3-\frac{2}{2x+1}\Big)$

$5)\; \; y=\sqrt{arcsin (\frac{3}{x}+2sin(x^5))}\\\\y'=\dfrac{1}{2\sqrt{arcsin (\frac{3}{x}+2sin(x^5))}}\cdot \dfrac{1}{\sqrt{1-\Big(\frac{3}{x}+2sin(x^5)\Big)^2}}\cdot \Big (-\dfrac{3}{x^2}+2\, cos(x^5)\cdot 5x^4\Big)$