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