anonymous
  • anonymous
Differentiate (2*n*b^(r*x)+n)^k. In this question when it asks to differentiate, with respect to which variable should we differentiate this function?
Mathematics
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katieb
  • katieb
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anonymous
  • anonymous
\[(2nb^{rx}+n)^k\] is the equation that has to be differentiated
amistre64
  • amistre64
you mean find a derivative?
anonymous
  • anonymous
yes

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amistre64
  • amistre64
if so; assume they are all functions of the same variable; and sort it out when you know more info :)
amistre64
  • amistre64
of course i got no idea how to recall to do an x^x type deal; but if you can remember it :) otherwise ill assume k to be constant \[D(2nb^{rx}+n)^k=k(2nb^{rx}+n)^{(k-1)}*D(2nb^{rx}+n)\] \[=k(2nb^{rx}+n)^{(k-1)}*(D(2nb^{rx})+D(n))\] \[=k(2nb^{rx}+n)^{(k-1)}*(D(2nb^{rx})+n')\] \[=k(2nb^{rx}+n)^{(k-1)}*(D(2n)b^{rx}+2nD(b^{rx})+n')\] \[=k(2nb^{rx}+n)^{(k-1)}*(2n'b^{rx}+2nD(b^{rx})+n')\] something similar to that is what id assume
anonymous
  • anonymous
Basically you have taken the derivative with respect to k
amistre64
  • amistre64
y = x^x Ly = Lx^x Ly = x Lx y'/y = x' Lx+x L'x y'/y = Lx+x/x y'/y = Lx+1 y' = y(Lx+1) y' = x^x (Lx+1) y' = x^x Lx + x^x might be what I was forgetting
amistre64
  • amistre64
well, I kept k constant and took the derivative of what I could with respect to say: time
amistre64
  • amistre64
\[(2(n(t))(b(t))^{(r(t)x(t))}+n(t))^k\] somehting like this
amistre64
  • amistre64
but thats only becasue I have no idea what the question is in context really
anonymous
  • anonymous
The question just says differentiate. Thats it. Doesn't even ask with respect to which variable should the derivative be taken. Because the derivative changes with each variable
anonymous
  • anonymous
For example, if the derivative is taken with respect to x, the derivative of the function has a different answer... if it is taken with respect to n, the answer changes accordingly.
amistre64
  • amistre64
if we are to take partial derivaitves; meaning that all the variables are independant ... then yes
amistre64
  • amistre64
i was thinking more of an implicit version

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