Question

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

Answer 1

is it just sqrtx

Answer 2

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

Answer 3

i think it would be 2/sqrt(x)

Answer 4

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

Answer 5

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

Answer 6

we used the general formula for the derivative of

if f(x) = ax^n

f'(x) = an x^(n-1)

Answer 7

oh ok thanks

Answer 8

yw

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

can someone please demonstrate that

Answer 13

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

Answer 14

h(t)=-16t^2 +30t + 200

first find the derivative of -16 t^2
using the rule

Answer 15

-32t + 30t + 0??

Answer 16

check derivative of 30t

Answer 17

30t is already reduce, isnt it?

Answer 18

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)

Answer 19

remember t^0 is 1

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

just 30

Answer 26

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

Answer 27

-32t + 30 + 0

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

ok i did and i got -34

Answer 32

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?

Answer 33

down

Answer 34

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

Answer 35

yeah it really did. thanks for taking the time.

Answer 36

to teach me