# Math Help - The fundamental theorem of calculus and the exponential function

1. ## The fundamental theorem of calculus and the exponential function

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...

2. Originally Posted by ALavoisier
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 $y = e^x$ then $\frac{dy}{dx} = e^x$.

Notice that $a^x = e^{\ln{a^x}}$

$= e^{x\ln{a}}$

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

So if $y = a^x$

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

Let $u = e^x$ so that $y = u^{\ln{a}}$.

$\frac{du}{dx} = e^x$

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

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

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

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

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

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

Therefore $\frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$

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

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

From here it is relatively easy to see that

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

Now, supposing you wanted to find

$\int{a^x\,dx}$

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

$= \frac{1}{\ln{a}}a^x + C$

$= \frac{a^x}{\ln{a}} + C$.

3. Thank you for your response. I was actually trying to take the derivative of that result using the product rule and chain rule or the quotient rule and get back the exponential function. Can you show me how

$\frac{d}{dx}(\frac{a^x}{\ln a})$ is equal to $a^x$ ?

4. Originally Posted by ALavoisier
Thank you for your response. I was actually trying to take the derivative of that result using the product rule and chain rule or the quotient rule and get back the exponential function. Can you show me how

$\frac{d}{dx}(\frac{a^x}{ln a})$ is equal to $a^x$ ?
$\frac{d}{dx}\left(\frac{a^x}{\ln{a}}\right) = \frac{1}{\ln{a}}\,\frac{d}{dx}(a^x)$.

The result follows.

5. Oh I am simply lending credence to the message in your sig...I wasn't treating the 1 over ln a as a constant, instead treating it as a natural log of a variable...embarrassing. Nonetheless it wasn't all for naught as I got to see your wonderful proof. Thank you for your time.