- anonymous

find the derivative of g(x)= 4sqrt(x)+3

- chestercat

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- anonymous

is it just sqrtx

- lalaly

\[\large{g'(x) =4 \times \frac{1}{2} x^{\frac{-1}{2}}} = \frac{2}{\sqrt{x}}\]

- anonymous

i think it would be 2/sqrt(x)

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## More answers

- anonymous

first convert the square root to a fractional index
= 4x^(1/2) + 3
derivative = 4 * (1/2) x ^(1/2 - 1) = 2 x(-1/2) or 2 / sqrtx

- anonymous

why did you have (1/2-1)...where did the -1 come from?

- anonymous

we used the general formula for the derivative of
if f(x) = ax^n
f'(x) = an x^(n-1)

- anonymous

oh ok thanks

- anonymous

yw

- anonymous

jimmy could you help me with this one...its confusing.
A girl standing on the top of a building 200 feet highs throws a water balloon up in the air at a velocity of 30ft/sec. The equation for the height of the water balloon is h(t)=-16t^2 +30t + 200
find the instantaneous velocity at two seconds and the direction of the water balloon

- phi

instantaneous velocity is dh/dt take the derivative with respect to t to get a new equation
then sub in t=2 to find the velocity

- phi

if you get +, the balloon is going higher. if you get a negative number, the balloon is falling

- anonymous

can someone please demonstrate that

- phi

find the derivative term by term
use if f(t) = a*t^n
f'(t) = a*n* t^(n-1)
just like Jim did up above

- phi

h(t)=-16t^2 +30t + 200
first find the derivative of -16 t^2
using the rule

- anonymous

-32t + 30t + 0??

- phi

check derivative of 30t

- anonymous

30t is already reduce, isnt it?

- phi

you start with
-16t^2 +30t + 200
then term by term
2* -16 * t^ (2-1)
30*1* t^(1-1)
200 * 0 * t(0-1)

- phi

remember t^0 is 1

- anonymous

oh i see so when i have -32t + 1 + 0...do i just plug in 2 to find the instantaneous velocity into t

- phi

of course nobody uses the exponent rule to find the derivative of a constant. We know it is zero.

- phi

now recheck : derivative with respect to t of : 30 t = ??

- anonymous

t^0= 1 therefore, 30t = 1?

- phi

use the rule: d/dt (30 t^1) = 1*30*t^(1-1) = 1*30*1 = ??

- anonymous

just 30

- phi

so the derivative of
-16t^2 +30t + 200 = ?

- anonymous

-32t + 30 + 0

- phi

adding 0 is not usually done . I mean leave it off, because it does not change anything

- phi

so now you have
h'(t) = -32t + 30
change in height with respect to time (velocity for this problem= -32t + 30

- phi

find the instantaneous velocity at two seconds and the direction of the water balloon
this means replace t with 2 in your equation to find the velocity at time t=2

- anonymous

ok i did and i got -34

- phi

you got -34 ft/sec (don't forget the units)
the negative sign means which direction?
Here is how we know which direction:
A girl throws a water balloon *up* in the air at a velocity of +30ft/sec.
The problem says +30ft/sec is up, so a -34 ft/sec must be which direction?

- anonymous

down

- phi

I hope it makes sense. But you have answered the entire question.

- anonymous

yeah it really did. thanks for taking the time.

- anonymous

to teach me

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