Question

Помогите решить задания
1 в и г
И второе задание

Answer 1

$1c)\ \ \lim\limits_{x \to 0}\, \dfrac{1-cos4x}{2x\cdot tg2x}=\lim\limits_{x \to 0}\, \dfrac{2sin^22x}{2x\cdot tg2x}=\Big[\ sin2x\sim 2x\ ,\ tg2x\sim 2x\ ,\ 2x\to 0\ \Big]=\\\\\\=\lim\limits_{x \to 0}\, \dfrac{2\cdot (2x)^2}{2x\cdot 2x}=\lim\limits_{x \to 0}\, \dfrac{8x^2}{4x^2}=\dfrac{8}{4}=2$

$1d)\ \ \lim\limits_{x \to \infty}\, \Big(\dfrac{3x+4}{3x+5}\Big)^{7x}=\lim\limits_{x \to \infty}\, \left(\Big(1+\dfrac{-1}{3x+5}\Big)^{\frac{3x+5}{-1}}\right)^{\frac{-7x}{3x+5}}=e^{\lim\limits_{x \to \infty}\, \frac{-7x}{3x+5}}=e^{-\frac{7}{3}}$

$2a)\ \ y=\dfrac{2x-3}{\sqrt{x^2+4x-3}}\\\\\\\dfrac{dy}{dx}=\dfrac{2\cdot \sqrt{x^2+4x-3}-(2x-3)\cdot \dfrac{2x+4}{2\sqrt{x^2+4x-3}}}{x^2+4x-3}=\\\\\\=\dfrac{2\cdot (x^2+4x-3)-(2x-3)\cdot (x+2)}{x^2+4x-3}=\dfrac{2x^2+8x-6-(2x^2+x-6)}{x^2+4x-3}=\\\\\\=\dfrac{7x}{x^2+4x-3}$

$2b)\ \ y=3^{cosx}\cdot arctg4x\ \ ,\ \ \ \ (uv)'=u'v+uv'\\\\y'=3^{cosx}\cdot ln3\cdot (-sinx)\cdot arctg4x+3^{cosx}\cdot \dfrac{4}{1+16x^2}\\\\\\2c)\ \ y=3^{arctgx}\cdot ln(1+4x^2)\\\\y'=3^{arctgx}\cdot ln3\cdot \dfrac{1}{1+x^2}\cdot ln(1+4x^2)+3^{arctgx}\cdot \dfrac{8x}{1+4x^2}$

$2d)\ \ y=(tg7x)^{x^3}\ \ ,\ \ \ y=e^{ln(tg7x)^{x^3}}\\\\y'=e^{ln(tg7x)^{x^3}}\cdot \Big(ln(tg7x)^{x^3}\Big)'=e^{ln(tg7x)^{x^3}}\cdot \Big(x^3\cdot ln(tg7x)\Big)'=\\\\\\=(tg7x)^{x^3}\cdot \left (3x^2\cdot ln(tg7x)+x^3\cdot \dfrac{1}{tg7x}\cdot \dfrac{7}{cos^27x}\right )$