Question

НАЙТИ ПРОИЗВОДНУЮ ФУНКЦИИ

Answer 1

Объяснение:

$3)\; \; f(x)=\sqrt{11x-x^2}\; \; \; ,\; \; \; (\sqrt{u})'=\frac{1}{2\sqrt{u}}\cdot u'\\\\f'(x)=\frac{1}{2\sqrt{11x-x^2}}\cdot (11-2x)\\\\\\4)\; \; f(x)=(2x-x^3)\sqrt{2-x^2}\; \; ,\; \; \; (uv)'=u'v+uv'\\\\f'(x)=(2-3x^2)\sqrt{2-x^2}+(2x-x^3)\cdot \frac{1}{2\sqrt{2-x^2}}\cdot (-2x)=\\\\=(2-3x^2)\sqrt{2-x^2}-\frac{x^2(2-x^2)}{\sqrt{2-x^2}}=(2-3x^2)\sqrt{2-x^2}-x^2\sqrt{2-x^2}=\\\\=\sqrt{2-x^2}\cdot (2-3x^2-x^2)=\sqrt{2-3x^2}\cdot (2-4x^2)\\\\\\5)\; \; f(x)=6cos^3(4-3x)\; \; ,\; \; \; (u^3)'=3u^2\cdot u'\\\\f'(x)=6\cdot 3cos^2(4-3x)\cdot (-sin(4-3x))\cdot (-3)=\\\\=54\cdot cos^2(4-3x)\cdot sin(4-3x)$

$6)\; \; f(x)=sin(4-3x)\cdot tg(4-3x)\; \; ,\; \; \; (uv)'=u'v+uv'\\\\f'(x)=cos(4-3x)\cdot (-3)\cdot tg(4-3x)+sin(4-3x)\cdot \frac{1}{cos^2(4-3x)}\cdot (-3)=\\\\=-3\cdot sin(4-3x)-\frac{3\, sin(4-3x)}{cos^2(4-3x)}=-3\cdot sin(4-3x)-\frac{3\, tg(4-3x)}{cos(4-3x)}=\\\\=3\cdot \frac{0,5\, sin(6x-8)+tg(3x-4)}{cos(3x-4)}$