Question

Найдите производную функции: 1)y=x^3-2x^2+x+2 2)y=√x(2sinx+1) 3)y=1/x^2 4)y=1/cosx 5)y=3x^2-2/x^3 6)y=tgx+1/x

Answer 1

$y=x^3-2x^2+x+2 \\\ y'=3x^2-2\cdot 2x+1=3x^2-4x+1$

$y= \sqrt{x} (2\sin x+1) \\\ y'=( \sqrt{x})' (2\sin x+1)+ \sqrt{x} (2\sin x+1)'= \\\ = \dfrac{1}{2 \sqrt{x} } (2\sin x+1)+ \sqrt{x} \cdot 2\cos x= \dfrac{\sin x}{ \sqrt{x} } + \dfrac{1}{2 \sqrt{x} } + 2\sqrt{x} \cos x$

$y= \dfrac{1}{x^2} =x^{-2} \\\ y'=-2x^{-2-1}=-2x^{-3}=- \dfrac{2}{x^3}$

$y= \dfrac{1}{\cos x} =(\cos x)^{-1} \\\ y'=-(\cos x)^{-1-1}\cdot (\cos x)'=-(\cos x)^{-2}\cdot (-\sin x)= \dfrac{\sin x}{\cos ^2x}$

$y=3x^2- \dfrac{2}{x^3} =3x^2- 2x^{-3} \\\ y'=3\cdot 2x- 2\cdot(-3x^{-4})=6x+ 6x^{-4}=6x+ \dfrac{6}{x^4}$

$y=\mathrm{tg}x+ \dfrac{1}{x} \\\ y'= \dfrac{1}{\cos^2x}- \dfrac{1}{x^2}$