how do you solve for the derivative of the square root of x+4?

Question

whats the derivative of this and can you show me how
to get to the answer

Answer 1

rewrite it as (x+4)^(1/2) then apply the power rule to it and get 1/2 * (x+4)^(-1/2) or 1/(2*sqroot(x+4))

Answer 2

The square root of x+4 is (x+4)^(.5), so you can take the derivative using the chain rule. It's (.5)*(x+4)^(-.5)

Answer 3

f (x) = (x + 4)^(1/2)
f `(x) = (1/2).(x + 4)^(-1/2)
f `(x) = 1 / [ 2.(x + 4)^(1/2) ]

Answer 4

f (x) = (2+8x)^(a million/2) f ' (x) = (a million/2)(2+8x)^(-a million/2)(8) f ' (x) = 4(2+8x)^(-a million/2) Rule: If f(x)= (a + bx)^d then f ' (x)= d(a+bx)^(d-a million)(b)...multiply the expression via the potential...and via the by-manufactured from the component interior the brackets...then subtract a million from the potential... area of f(x)... because of the fact it incredibly is below a sq. root...2+8x would desire to be effective... so... 2+8x > 0 8x > -2 x > -a million/4

Answer 5

let y = Sqrt(x+4)
=> y = (x+4)^1/2

Let u = (x+4), => y = u^1/2
du/dx = 1, => dy/du = (1/2) u^(-1/2)

By applying Chain Rule,
dy/dx = dy/du * du/dx
= (1/2) u^(-1/2) * 1
= (1/2) u^(-1/2)
=> = (1/2) (x+4)^(-1/2)
= 1/ [2*sqrt(x+4)]

Answer 6

By applying Chain Rule,
dy/dx = dy/du * du/dx
= (½) u^(-½)(1)
= (½) u^(-½)
= (½) (x+4)^(-½)
= 1/ [2(x+4)(-½)]
= 1/ [2√(x+4)]

Answer 7

d/dx g(f(x)) = dg/df * df/dx In this problem, f(x) = x+4, and g(f) = √f

d/df √f = 1/2(f^-1/2) = (1/2)(x+4)^-1/2

dg/dx = d(x+4)/dx = 1

The product is then (1/2)(x+4)^-1/2