Question

Найти производные функций.

Answer 1

Ответ:

$a)\ \ y=\dfrac{1}{3}\, x^3\cdot tgx+ln\, cos\sqrt{x}+e^{5x}\\\\y'=x^2\cdot tgx+\dfrac{1}{3}\, x^3\cdot \dfrac{1}{cos^2x}+\dfrac{1}{cos\sqrt{x}}\cdot (-sin\sqrt{x})\cdot \dfrac{1}{2\sqrt{x}}+5e^{5x}=\\\\=x^2\cdot tgx+\dfrac{x^3}{3\cdot cos^2x}-\dfrac{tg\sqrt{x}}{2\sqrt{x}}+5e^{5x}$

$b)\ \ y=ln\sqrt{\dfrac{x^2+1}{x+1}}\\\\y'=\sqrt{\dfrac{x+1}{x^2+1}}\cdot \dfrac{1}{2\sqrt{\dfrac{x^2+1}{x+1}}}\cdot \dfrac{2x(x+1)-(x^2+1)}{(x+1)^2}=\dfrac{x+1}{2(x^2+1)}\cdot \dfrac{x^2+2x-1}{(x+1)^2}=\\\\\\=\dfrac{x^2+2x-1}{2(x^2+1)(x+1)}$

$c)\ \ x^3y^3-2xy+3x=0\\\\(x^3)'\cdot y^3+x^3\cdot (y^3)'-(2x)'\cdot y-2x\cdot y'+(3x)'=0\\\\3x^2\cdot y^3+x^3\cdot 3y^2y'-2y-2xy'+3=0\\\\y'\cdot (3x^3y^2-2x)=2y-3x^2y^3-3\\\\y'=\dfrac{2y-3x^2y^3-3}{3x^3y^2-2x}$