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bronzegoddess
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differentiate the following, I am stuck so I really don't know what to do..
 one year ago
 one year ago
bronzegoddess Group Title
differentiate the following, I am stuck so I really don't know what to do..
 one year ago
 one year ago

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bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
\[f(N)= \frac{ RN }{ (1+aN)^b }\]
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
a, b and R are positive constants
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
use the quotient rule
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
ok so we get: \[\frac{'( RN) (1+aN)^b  (RN)'(1+aN)^b}{ \left[ (1+aN)^b \right]^2}\]
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
is that it? my book says we also have to use the chain rule but am so confused...
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
you have to use the chain rule to differentiate (1 + aN)^b derivative of this is ab((1 + aN)^(b1)
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
this is the answer: \[\frac{ R }{ (1+aN)^{b+1}}(1abN+aN)\]
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
hold on  i'll check that out on paper
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
ok thank you! :)
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
yea thats right but its a bit messy to set it out  i'll try the draw facility
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
ok, i dnt mind a scan...
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
its not working = R(1 + aN)^b  RNab(a + aN)^(b1)  (1 + aN)^2b = R(1 + aN)^(b1) ( ((1 + aN)  abN)  (1 + aN)^2b = R(1 + aN  abN)  (1 + aN)^(b+1)
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
ok thank you! ^^
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
wait, can i ask a few questions to see if i understand?
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
i dnt understand how you get this, i know you said chain rule? ab(a + aN)^(b1)
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
if we have a function of a function like (1 + 2x)^3 the derivative of this is found this way: (using the dy/dx notation): let u = 1 + 2x , then y = u^3 so dy/du = 3u^2 and du/dx = 2 so dy/dx = dy/du * du/dx = 2 * 3 u^2 = 6 u^2 = 6(1 + 2x)^2 you can use the notation f'(x) to show the same thing but personally i prefer the 'd' notation. ithe differentiation in your question can be done similarly
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
thats the chain rule or 'function of a function' rule
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
oh its easier with the Liebniz notation! thank you sooo much! :)
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
the derivative of the 'inside function' (1 + aN) = a and if we let (1 + aN)^b = u^b its derivative is bu^(b1)
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
yep  i find the Leibnitz notation easier for the chain rule
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
my problem is noticing the chain rule because to be the (1+aN)^b didnt look like one :/
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
right maybe its having the lettered constants there intread of actual numbers
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
yes :( i hate applied math, i have an exam tomorrow at 2pm =.=
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
look out for a function within a function
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
i will :)
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.1
 that was a tricky one and the algebra was a bit bit messy too good luck
 one year ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
thank you!^^
 one year ago
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