I was just fooling around today seeing how derivatives and integrals go back and forth and i tried to take the derivative of the result of the integral of the exponential function (a raised to the x). I.E. i took the derivative of a to the x divided by ln a in the hopes that i would get back the exponential function. I seem to get something slightly different. Am I failing in my algebraic simplification or does this have something to do with the constant of integration that gets stamped on the result of the indefinite integral of the exponential function ( a to the x over ln a + C). I am fairly new to calculus and my previous algebra and trig was a few year ago...

Oh and I promise to learn to post in the "math font" soon...

By definition, if \(\displaystyle y = e^x\) then \(\displaystyle \frac{dy}{dx} = e^x\).

Notice that \(\displaystyle a^x = e^{\ln{a^x}}\)

\(\displaystyle = e^{x\ln{a}}\)

\(\displaystyle = (e^x)^{\ln{a}}\).

So if \(\displaystyle y = a^x\)

\(\displaystyle y = (e^x)^{\ln{a}}\).

Let \(\displaystyle u = e^x\) so that \(\displaystyle y = u^{\ln{a}}\).

\(\displaystyle \frac{du}{dx} = e^x\)

\(\displaystyle \frac{dy}{du} = (\ln{a})u^{\ln{a} - 1}\)

\(\displaystyle = (\ln{a})(e^x)^{\ln{a} - 1}\)

\(\displaystyle = (\ln{a})(e^{x\ln{a} - x})\)

\(\displaystyle = (\ln{a})(e^{x\ln{a}})(e^{-x})\)

\(\displaystyle = (\ln{a})(e^{\ln{a^x}})(e^{-x})\)

\(\displaystyle = (\ln{a})(a^x)(e^{-x})\).

Therefore \(\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}\)

\(\displaystyle = e^x(\ln{a})(a^x)(e^{-x})\)

\(\displaystyle = (\ln{a})a^x\).

From here it is relatively easy to see that

\(\displaystyle \int{(\ln{a})a^x\,dx} = a^x + C\).

Now, supposing you wanted to find

\(\displaystyle \int{a^x\,dx}\)

\(\displaystyle \int{a^x\,dx} = \frac{1}{\ln{a}}\int{(\ln{a})a^x\,dx}\)

\(\displaystyle = \frac{1}{\ln{a}}a^x + C\)

\(\displaystyle = \frac{a^x}{\ln{a}} + C\).

Hope this answered your question.