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bronzegoddess
differentiate the following, I am stuck so I really don't know what to do..
\[f(N)= \frac{ RN }{ (1+aN)^b }\]
a, b and R are positive constants
ok so we get: \[\frac{'( RN) (1+aN)^b - (RN)'(1+aN)^b}{ \left[ (1+aN)^b \right]^2}\]
is that it? my book says we also have to use the chain rule but am so confused...
you have to use the chain rule to differentiate (1 + aN)^b derivative of this is ab((1 + aN)^(b-1)
this is the answer: \[\frac{ R }{ (1+aN)^{b+1}}(1-abN+aN)\]
hold on - i'll check that out on paper
yea thats right but its a bit messy to set it out - i'll try the draw facility
ok, i dnt mind a scan...
its not working = R(1 + aN)^b - RNab(a + aN)^(b-1) ------------------------------- (1 + aN)^2b = R(1 + aN)^(b-1) ( ((1 + aN) - abN) ----------------------------- (1 + aN)^2b = R(1 + aN - abN) --------------- (1 + aN)^(b+1)
wait, can i ask a few questions to see if i understand?
i dnt understand how you get this, i know you said chain rule? ab(a + aN)^(b-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
thats the chain rule or 'function of a function' rule
oh its easier with the Liebniz notation! thank you sooo much! :)
the derivative of the 'inside function' (1 + aN) = a and if we let (1 + aN)^b = u^b its derivative is bu^(b-1)
yep - i find the Leibnitz notation easier for the chain rule
my problem is noticing the chain rule because to be the (1+aN)^b didnt look like one :/
right maybe its having the lettered constants there intread of actual numbers
yes :( i hate applied math, i have an exam tomorrow at 2pm =.=
look out for a function within a function
- that was a tricky one and the algebra was a bit bit messy too good luck