d/dx(a^x)=?

[ahorn]

Hello

My question is what d/dx(a^x) equals. My answer is $\displaystyle a^x \cdot \lim_{h \to 0} \frac{a^h-1}{h}$ but is that the simplest form?

 

[Quaid]

My question is what d/dx(a^x) equals.

My answer is $\displaystyle a^x \cdot \lim_{h \to 0} \frac{a^h-1}{h}$ but is that the simplest form?

Hi ahorn:

At the introductory level, that's a very good start.

Calculating that limit is generally beyond the scope of a beginning course in derivatives. (Your text ought to provide it.)

$\displaystyle \lim\limits_{h \to 0} \dfrac{a^h - 1}{h} = ln(a)$

Now, what happens when a = e?

Cheers :cool:

 

[HallsofIvy]

One way of doing this to note that, since $\displaystyle e^x$ and $\displaystyle ln(x)$ are "inverse functions", $\displaystyle a^x= e^{ln(a^x)}= e^{x ln(a)}$. Now, use the chain rule, $\displaystyle \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}$ with $\displaystyle u= xln(a)$.

 

[ahorn]

@Quaid: e^x differentiated = e^x


@Hallsofivy: Differentiation using the chain rule:

\(\displaystyle \frac{d}{dx}a^x\\

= \frac{d}{dx}e^{\ln(a^x)}\\

= \frac{d}{dx}e^{x \ln(a)}\\

= e^{x \ln(a)} \cdot (ln(a) + x/a) \\

= a^x \ln(a) + x \cdot a^{x-1}\)


I almost got $\displaystyle a^x \ln(a)$ by itself... can you see where I went wrong?

 

[ahorn]

Oh. I differentiated x*ln(a) incorrectly. ln(a) is a constant.

 

[HallsofIvy]

Oh. I differentiated x*ln(a) incorrectly. ln(a) is a constant.

Yes, that was your error. The derivative of $\displaystyle a^x$ is $\displaystyle a^x ln(a)$. Note that ln(e)= 1 so that if a= e, that becomes $\displaystyle \frac{d e^x}{dx}= e^x(1)= e^x$.