Question

Помогите решить задачу :Нужно найти производную.
От меня лайк и подписка. ​

Answer 1

$1)\; \; y=\frac{x^3}{tg^22x}\\\\y'=\frac{3x^2\cdot tg^22x-x^3\cdot 2\, tg2x\cdot \frac{1}{cos^22x}\cdot 2}{tg^42x}=\frac{3x^2\cdot tg2x-4x^3\cdot \frac{1}{cos^22x}}{tg^32x}=\\\\=\frac{cos2x\cdot (3x^2\cdot cos2x\cdot sin2x-4x^3)}{sin^32x}=\frac{x^2\cdot cos2x\cdot (1,5\cdot sin4x-4x)}{sin^32x}$

$2)\; \; y=\frac{arctg\frac{2x}{3}}{4x^2+9}\\\\y'=\frac{1}{(4x^2+9)^2}\cdot \Big (\frac{1}{1+\frac{4x^2}{9}}\cdot \frac{2}{3}\cdot (4x^2+9)-arctg\frac{2x}{3}\cdot 8x\Big )=\\\\=\frac{1}{(4x^2+9)^2}\cdot \Big (6-8x\cdot arctg\frac{2x}{3}\Big )$

$3)\; \; siny+x^2y=1\\\\cosy\cdot y'+2x\cdot y+x^2\cdot y'=0\\\\y'\cdot (cosy+x^2)=-2xy\\\\y'=-\frac{2xy}{cosy+x^2}\\\\\\4)\; \; \left \{ {{x=arcsint} \atop {y=\sqrt{1-t^2}}} \right.\\\\x'_{t}=\frac{1}{\sqrt{1-t^2}}\; \; ,\; \; \; y'_t}=\frac{-2t}{2\sqrt{1-t^2}}=-\frac{t}{\sqrt{1-t^2}}\\\\y'_{x}=\frac{y'_{t}}{x'_{t}}=-\frac{t}{\sqrt{1-t^2}}:\frac{1}{\sqrt{1-t^2}}=-t=-sinx$