If f(x)= sqrt(4sin(x)+2), then f'(0)=

Answer 1

Use the substitution
u=4sin(x)+2
f'(x)=df(x)/dx
=df(u)/du * du/dx
=d(sqrt(u))/du * d(4sin(x)+2)/dx
=1/(2sqrt(u)) * 4cos(x)
=2cos(x)/sqrt(4sin(x)+2))

Answer 2

Well the Sqrt(4sin(x)+2) is the same thing as saying (4sin(x)+2)^(1/2)....sooo take the derivative using the chain rule. So first you would take the derivative of the outside by bringing the (1/2) and subtracting one from the exponent. Then take the derivative of the inside. 4 is a constant so it remains the same since it being multiplied. Sin(x)has the derivative cos(x) and since the two is a constant its derivative is zero. Your answer should look like this:
(4cos(x))/(2sqrt(4sin(x)+2)

Answer 3

radical 2

Answer 4

rad(4sinx +2) or (4sinx +2)^1/2
chain rule
1/2(4sinx +2)^-(1/2) * 4cosx <----- derivative of sinx is cosx
substitue x for 0 and plug into calulator as follows:
1/2 (4sin(0) +2) ^-(1/2) *4cos(0)
final answer
f'(0) aprox. 1.414 or radical(2), squar root(2) same thing