Question

Найти производную(распишете подробно): 1) y= - 3cosx * (x`2 + 2) 2)y=1/на корень из x 3) y= 1/sinx 4)y=x`4/ 3-x

Answer 1

1)
$y' = (- 3cosx * ( x^{2} +2))' = -3 [(cosx)' * ( x^{2} +2)+cosx * (x^{2} +2)'] = \\ \\ = -3 [-sinx * ( x^{2} +2)+cosx * 2x] = -3 (2x*cosx-(x^{2} +2)*sinx)$

2) 
$y' = ( \frac{1}{ \sqrt{x} })' = ( {x}^{-1/2} })' = - \frac{1}{2} * x^{-3/2} = - \frac{1}{2 \sqrt{x^3} }$

3) 
$y' = ( \frac{1}{sinx} )' = (sin^{-1}x)' = - 1 *sin^{-2}x*(sinx)' = - \frac{cosx}{sin^2x}$

4) 
$y' = ( \frac{x^4}{3-x})' = \frac{(x^4)'(3-x)-x^4(3-x)'}{(3-x)^2} = \frac{4x^3(3-x)-x^4(-1)}{(3-x)^2} = \\ \\ = \frac{12x^3-4x^4+x^4}{(3-x)^2} = \frac{12x^3-3x^4}{(3-x)^2} = \frac{3x^3(4-x)}{(3-x)^2}$