f(x)=(3x^4)*sqrt(x) + 7/x^2*(sqrt(x))
f '(x)= ??

- anonymous

f(x)=(3x^4)*sqrt(x) + 7/x^2*(sqrt(x))
f '(x)= ??

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- amistre64

change the square root to an exponent

- amistre64

3x^4 x^(1/2) + 7/(x^2 x^(1/2))

- amistre64

3x^(9/2) + 7/(x^5/2) and work that....

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## More answers

- amistre64

do you know how to do a derivative?

- amistre64

7/(x^5/2) is its own term right? or is that (x^(5/2)) the demoniator of the whole equation?

- amistre64

d..e...n..o..m...i..n...a..t...o....r.. lol

- anonymous

own term

- anonymous

haha

- amistre64

good... :)

- anonymous

o shoot , i thought i was replying to you but it wasnt posting

- anonymous

yes, i know how to find derivatives

- anonymous

i just dont know where i am going wrong

- amistre64

step me thru what you have done and I will let you know :)

- anonymous

(i had made them exponents as well)

- amistre64

the first term is a breeze.... just multiply exponent and subtract "1" from it.....

- amistre64

3x^(9/2) = 3(9/2)x^(7/2) =
27 x^(7/2)
---------
2
right?

- anonymous

sorry got kicked out, yes that is what i have

- amistre64

x^(7/2) = x^(6/2) x^(1/2) so if you want to clean it up some:
17 x^3 sqrt(x)
------------ for the first term...
2

- amistre64

7
-------- how would you derive this term?
(x^(5/2))

- anonymous

ok

- anonymous

oo i had x^3/2 not x^5/2

- amistre64

quotient rule:
I remember it like this:
BT'-B'T
------ where B=bottom, and T=top
B^2

- amistre64

whose right then, me or you :)

- anonymous

when you are combining x^2 and x^1/2 dont you add the exponents?

- amistre64

yes... 2 + 1/2 = a mixed numeral...tern it into an improper fraction
2(2)+1 5
------ = ---
2 2

- amistre64

lol...a tern is a sea bird i think

- amistre64

7
-------- lets derive this...
(x^(5/2))

- anonymous

oo duh.

- anonymous

ok simple addition.

- anonymous

so that would be--

- anonymous

-35/2x^7/2

- anonymous

may i ask you one more derivative question?

- amistre64

5 x^(3/2)
B' = -------- and T' = 0
2
-35 x^(3/2)
--------------
2
------------------
x^5
Is what I get...

- amistre64

sure, but lets finish this one first :)

- anonymous

i plugged it in my school system and got the right answer :)

- amistre64

lol ... cheater :)

- amistre64

next question?

- anonymous

they dont tell you what the answer is, just if you get it correct ;)

- anonymous

if f(x) = sqrt(6x)
f 'x = ?

- amistre64

2 ways to do this; we can do a substituion and the chain rule? or simply change to exponent and solve it.. which way you want it?

- anonymous

change to exponent

- amistre64

ok....
sqrt(6x) = (6x)^(1/2) ....might have to still do the "extra" because of that (6x) in there...

- anonymous

ok

- anonymous

is this correct-- 1/(2(6x)^1/2)

- amistre64

(1/2)(6x)^(-1/2) cleans up to:
1
-------- ;we got to multiply (use the chain rule) to use that "6x"
2sqrt(6x)

- amistre64

that didnt turn out the way I wanted lol

- amistre64

1
------- times Dx(6x)
2sqrt(6x)

- amistre64

Dx(6x) = 6

- anonymous

i dont understand why you are taking the derivative of 6x

- amistre64

6/2 = 3
3
-------
sqrt(6x)

- amistre64

let me explain it like this:
remember when we have f(g(x)) and want to derive it?

- amistre64

we have to use the "chain rule" to solve it right?
this problem is exactly that set up, let me show you..

- anonymous

sorry, keep getting kicked out. i havent learned the chain rule

- amistre64

y = sqrt(u) and u = 6x
to solve dy/dx we need the chain rule:
(dy/dx) = (dy/du) (du/dx)
y = sqrt(u)
dy/du = 1/2sqrt(u)
u = 6x
du/dx = 6
Do you see it?

- anonymous

yes

- amistre64

Do you understand it?
Imagine a set of gears that are meshed together, when the first one is turned, all the others turn as well. But they depend on the funtion of the first one to determine how they are going to behave.... does that makes sense?

- anonymous

ah, yes

- anonymous

now i understand how you got to your final answer. thank you so much for all of your help!!!

- amistre64

youre welcome :)

- anonymous

definitely a 'fan' of you :)

- amistre64

:) thanx...

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