Find the derivative: y = X^3 sqrt (X+2)?

Answer 1

first, rewrite:
y = x^3*(x+2)^(1/2)
now, use product rule:
dy/dx = fg' + f'g
f is x^3 and g is (x+2)^(1/2)
f' is 3x^2 and g' is (1/2)(x+2)^(-1/2) (we multiply by the exponent and the derivative of the inside, and decrease the exponent by 1)
so f'g = 3x^2*sqrt(x+2)
and fg' = (3x^2)/(2*sqrt(x+2))
so your answer is:
3x^2*sqrt(x+2) + (3x^2)/(2*sqrt(x+2))
pull out a 3x^2 and get:
(3x^2)(sqrt(x+2) + 1/(2*sqrt(x+2)))

Answer 2

y= x^3(x+2)^1/2
diff. w.r.t x,we get
dy/dx= 3x^2(x+2)^1/2+x^3*1/2*1/(x+2)^1/2 *1
= 3/2 x^2(x+2)^1/2+1/2 x^3(x+2)^(-1/2)

Answer 3

dy / dx = 3x².(x + 2)^(1/2) + (1 / 2).(x + 2)^(- 1/2).x³
dy / dx = x².(x + 2)^(- 1/2)[ 3(x + 2) + (1/2).x ]
dy / dx = x².(x + 2)^(- 1/2)[ (7 / 2).x + 12 / 2]
dy / dx = (1 / 2).(7x + 12).x² / √ [ x + 2]