## anonymous 4 years ago find the derivative of g(x)= 4sqrt(x)+3

1. anonymous

is it just sqrtx

2. lalaly

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

3. anonymous

i think it would be 2/sqrt(x)

4. 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

5. anonymous

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

6. anonymous

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

7. anonymous

oh ok thanks

8. anonymous

yw

9. 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

10. 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

11. phi

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

12. anonymous

13. 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

14. phi

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

15. anonymous

-32t + 30t + 0??

16. phi

check derivative of 30t

17. anonymous

30t is already reduce, isnt it?

18. 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)

19. phi

remember t^0 is 1

20. anonymous

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

21. phi

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

22. phi

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

23. anonymous

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

24. phi

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

25. anonymous

just 30

26. phi

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

27. anonymous

-32t + 30 + 0

28. phi

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

29. phi

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

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

31. anonymous

ok i did and i got -34

32. 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?

33. anonymous

down

34. phi

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

35. anonymous

yeah it really did. thanks for taking the time.

36. anonymous

to teach me