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change the square root to an exponent

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

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

do you know how to do a derivative?

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

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

own term

haha

good... :)

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

yes, i know how to find derivatives

i just dont know where i am going wrong

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

(i had made them exponents as well)

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

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

sorry got kicked out, yes that is what i have

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

ok

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

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

whose right then, me or you :)

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

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

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

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

oo duh.

ok simple addition.

so that would be--

-35/2x^7/2

may i ask you one more derivative question?

sure, but lets finish this one first :)

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

lol ... cheater :)

next question?

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

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

change to exponent

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

ok

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

that didnt turn out the way I wanted lol

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

Dx(6x) = 6

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

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

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

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

yes

ah, yes

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

youre welcome :)

definitely a 'fan' of you :)

:) thanx...