Calculus Question, Need Help.. Find f''(x)?

Question

Find f''(x) if f(x) = √4x+9

(just so u know that is the square root of 4x+9)

______
V 4x+9

need solution please

you are all of a great help, i have done almost 107 calculus problems over the past 4 days and have been stumped on about 20 or so... Thank you so much for your time and help

Answer 1

Use the chain rule. If h(x) = f(g(x)), then h'(x) = f'(g(x))*g'(x). We call g(x) the inner function and f(x) the outer function. Here, the outer function is sqrt(x) or x^0.5 and the inner function is 4x + 9. The derivative of the outer function is 1 / (2*sqrt(x)) or 0.5x^-0.5 and the derivative of the inner function is 4. Just plug the inner function in for x in the derivative of the outer function, and multiply by the derivative of the inner function. I get 2 / sqrt(4x + 9) or 2(4x + 9)^-0.5. To get the second derivative of f, just take the derivative of the first derivative, again using the chain rule.

Answer 2

This is a chain rule question.
in general f(g(x)) --> f'(g(x)) * g'(x) dx

so for this equation, f is the square root and g is 4x + 9.

So you'd take the derivative of square root, which is 1/(2sqrt(x))
Just replace x with 4x + 9 to get f'(g(x)).

1 / (2sqrt(4x + 9))

then multiply that times the derivative of 4x + 9, which is just 4.
So altogether, you have

1 / (2sqrt(4x + 9)) * 4 =
4 / (2sqrt(4x + 9)) =
2 / sqrt(4x + 9)

(the sqrt = square root.)

Answer 3

well, the square root would mean that the 4x+9 is raised to the 1/2 power. You can then apply the chain rule.

Answer 4

hi.

the square root of 4x + 9 is the same as (4x +9)to the power of 1/2.


so f'(x) = (4x+9)1/2 **note:1/2 is an exponential power

then f''(x) = 1/2(4x+9)to the power of -1/2 multiplied by 4

then f''(x) = 2(4x+9)to the power of -1/2

final answer: f''(x) = 2 over square root of 4x +9

++note: its a fraction :)
hope it helps.

add me on your ym.
xxgreenslippersxx_05 thanks :))

Answer 5

this is a chain rule problem, rewriting the problem we get

f(x) = (4x+9)^(1/2)

so f'(x) = [(1/2)(4x+9)^(-1/2)](4)

f''(x) = 2[(-1/2)(4x+9)^(-3/2)](4)

you can reduce both answers

Answer 6

asdqd

Answer 7

F(X)= (4X + 9)^ (1/2)
F' = (0.5)*[(4X + 9) ^(- 1/2)] * (4)= 2 (4X + 9 )^ (-1/2)
F'' = 2(-1/2)[(4X+9)^(-3/2)](4)= - 4 [(4X+9)^(-3/2)]

Answer 8

(4x+9)^.5

f'(x) = 4/ (4x+9)^.5

f''(x) = 16/(4x+9)^1.5


I haven't done calc in a loong time so I don't remember if I did it right but I think I did...

Answer 9

let u = 4x + 9

f(x) = (u)^.5
f'(x) = .5 * (u)^(-.5) * du = .5(4x + 9)^-.5 * 4 = 2u^-.5
f''(x) = 2*-.5*u^(-1.5) * du = -(4x+9)^-1.5 *4

f''(x) = -4(4x+9)^(-1.5)

Answer 10

set u^2 = 4*x+9, f = u

df/dx = df/du*du/dx

2udu = 4dx , du/dx = 2/root(4x+9) = 2/u
df/dx = 2/u = 2/root(4x+9)

d2f/dx^2 = d/du ( df/dx) = d/du (df/dx) *du/dx

d2f/dx^2 = -2/u^3 * 2/u = -4/u^4 = -4/(4x+9)^2