I'm trying to find how to take the derivative of 2^x by use of the difference quotient, rather than simply using derivative rules. I know it's ln(2) * 2^x, but I need to essentially prove it. Does anyone know how?

- Oct 28th 2008, 01:32 AMpantsaregoodFinding the derivative of an exponential using the difference quotient
I'm trying to find how to take the derivative of 2^x by use of the difference quotient, rather than simply using derivative rules. I know it's ln(2) * 2^x, but I need to essentially prove it. Does anyone know how?

- Oct 28th 2008, 03:58 AMtoraj58
$\displaystyle u = 2^x$

so

$\displaystyle \ln(u) = \ln(2^x) $

differntiate two sides of the equation:

$\displaystyle (\ln(u))^\prime = (\ln(2^x))^\prime $

left is equat to : $\displaystyle \frac {u^\prime}{u} $

so we have:

$\displaystyle \frac {u^\prime}{u} = (x\ln 2)^\prime$

then we have:

$\displaystyle u^\prime = u(x\ln 2)^\prime$

$\displaystyle u^\prime = 2^x(1\times\ln 2 + 0\times x)$

$\displaystyle u^\prime = 2^x\ln 2$ - Oct 28th 2008, 05:24 AMHallsofIvy
But toraj58's response doesn't address the question: find the integral using the difference quotient; i.e. the basic definition of the derivative.

pantsaregood, the difference quotient for a function f if [f(x+h)- f(x)]/h. What is that when $\displaystyle f(x)= 2^x$? You can use one of the laws of exponents to simplify that a little. - Oct 28th 2008, 05:41 PMpantsaregood
2^x

[2^(x+h)-2^x]/h

That really doesn't get me anywhere, though.