2
$\begingroup$

I am tasked with taking the first and second derivative of a polynomial function. I chose Moore's law which states that every two years, the amount of transistors double. The equation for this law is:

$$ P_n = P_0\times2^{1/2n} $$

  • $P_n$ is the computer processing power n years after the current year.
  • $P_0$ is the computer processing power in the current year.
  • $n$ is the number of years after the current year.

My question is to take the derivative of this do I drop the $P_n$ because it is a constant? Or is this a matter of using Leibniz's notation vs Lagrange's notation? Should I be using implicit differentiation because I have more than one variable? I know the rules of differentiation I am just not sure of the format because I am used to seeing equations like $f(x)= ...$

Thank you for your help.

$\endgroup$
  • 6
    $\begingroup$ Moore's law is an exponential function, not a polynomial function $\endgroup$ – Thomas Andrews Jul 29 '17 at 22:47
  • $\begingroup$ Your formula for $P_n$ now has an $x$ in it, what does $x$ denote? Also even without the $x$ your formula makes it seems the number of transistors is divided by $2$ each year, not doubled. Or maybe you mean the power to be $(1/2)n.$ $\endgroup$ – coffeemath Jul 29 '17 at 22:54
  • $\begingroup$ @ThomasAndrews Ah yes, thank you for the correction. $\endgroup$ – Bobby Jul 29 '17 at 23:18
  • 1
    $\begingroup$ @coffeemath The x is denoting multiplication. And yes that was a formatting error. It should be 2^(n/2) as stated below by Lee Mosher. $\endgroup$ – Bobby Jul 29 '17 at 23:19
3
$\begingroup$

First, there is only one independent variable, and that is $n$. The $P_0$ is a constant. The $P_n$ is a dependent variable. This answers one of your questions:

  • $P_n$ is not a constant.

Second, your formula is ambiguous, and I suspect that you mean $$P_n = P_0 2^{(1/2)n} $$ which simplifies to $$P_n = P_0 2^{n/2} $$

Third, you seem to be thinking of $n$ as having discrete values, i.e. whole number values. However, if you instead think of the independent variable $n$ as having continuous, i.e. real number values, and if you therefore use $x$ instead as $n$, and if you also think of the dependent variable $P_n$ as having real number values, and if you therefore use $y$ instead of $P_n$, then you get $$y = P_0 2^{x/2} $$ And now, yes, you are perfectly justified in using the equation $y=f(x)$ and rewriting in function notation as $$f(x) = P_0 2^{x/2} $$

Fourth, this is not a polynomial function. If it were a polynomial function, the $x$ would be in the base of the exponential expression. Since $x$ is instead in the exponent of the exponential expression, so this is an exponential function. Taking the derivatives of exponential functions follows different rules than taking the derivatives of polynomial functions, as you may know. In particular, to answer another one of your questions:

  • The notation (Leibniz vs. Lagrange) does not matter so much as recognizing the correct type of function and applying the corresponding differentiation rules.

Fifth, and finally, you don't need to use implicit differentiation, if you know the rule for differentiating exponential functions. (On the other hand, one common derivation for the exponential function rule does indeed involve implicit differentiation...)

$\endgroup$
  • $\begingroup$ That makes so much more sense, thank you for your help! $\endgroup$ – Bobby Jul 29 '17 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.