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

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

Mathematics
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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....

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Other answers:

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
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
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?
5 x^(3/2) B' = -------- and T' = 0 2 -35 x^(3/2) -------------- 2 ------------------ x^5 Is what I get...
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 = ?
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?
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)
(1/2)(6x)^(-1/2) cleans up to: 1 -------- ;we got to multiply (use the chain rule) to use that "6x" 2sqrt(6x)
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?
we have to use the "chain rule" to solve it right? this problem is exactly that set up, let me show you..
sorry, keep getting kicked out. i havent learned the chain rule
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?
yes
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?
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...

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