Question

Помогите решить плиз..
4,6,8,12

Answer 1

y =  2/(x+1)
y` = - 2/(x+1)^2 * 1 = - 2/(x+1)^2

y=(x^5+1)^(1/2)
y` = 1/2 * 1/(x^5+1)^(1/2) * 5x^4 = (5x^4) / 2 *  1 / (x^5+1)^(1/2)

y=((2x^2-cos(x)+2)^3)^(1/4) = (2x^2-cos(x)+2)^(3/4)
y`= (3/4) * 1/(2x^2-cos(x)+2)^(1/4)  * ( 4x+sin(x) )

y=1/(ctg(x))^3 = (tg(x))^3
y`=3* (tg(x))^2 * 1/(cos(x))^2

Answer 2

Спасибо!