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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 2

Lesson 8: Derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x)

# Proof: The derivative of 𝑒ˣ is 𝑒ˣ

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.A (LO)
,
FUN‑3.A.4 (EK)
e, start superscript, x, end superscript is the only function that is the derivative of itself!
start fraction, d, divided by, d, x, end fraction, open bracket, e, start superscript, x, end superscript, close bracket, equals, e, start superscript, x, end superscript
(Well, actually, f, left parenthesis, x, right parenthesis, equals, 0 is also the derivative of itself, but it's not a very interesting function...)
The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.
Proof: The derivative of 𝑒ˣ is 𝑒ˣSee video transcript

## Want to join the conversation?

• At , how did the limit got inside the logarithm function? It is getting hard for me to make sense for this step. It is like saying lim (x -> 0) cos(x) = cos (lim x->0 x).
How is that possible?
Can this thing be only applied to logarithm functions or is it generic for other functions also like cos, sin etc?
• It's NOT a general rule, and I wish Sal spent some time explaining why it works in this particular case.

– – –

First of all, we're dealing with a composite function.

𝑓(𝑥) = 1∕ln 𝑥
𝑔(𝑥) = (1 + 𝑥)^(1∕𝑥)

𝑓(𝑔(𝑥)) = 1∕ln((1 + 𝑥)^(1∕𝑥))

In general terms we are looking for
𝐹 = lim(𝑛 → 0) 𝑓(𝑔(𝑛))

This means that we let 𝑛 approach zero, which makes 𝑔(𝑛) approach some limit 𝐺, which in turn makes 𝑓(𝑔(𝑛)) approach 𝐹.

In other words:
𝐺 = lim(𝑛 → 0) 𝑔(𝑛)
𝐹 = lim(𝑔(𝑛) → 𝐺) 𝑓(𝑔(𝑛)) = [let 𝑥 = 𝑔(𝑛)] = lim(𝑥 → 𝐺) 𝑓(𝑥)

Now, if we use our definitions of 𝑓(𝑥) and 𝑔(𝑥), we get
𝐺 = lim(𝑛 → 0) (1 + 𝑛)^(1∕𝑛) = [by definition] = 𝑒
𝐹 = lim(𝑥 → 𝑒) 1∕ln 𝑥 = [by direct substitution] = 1∕ln 𝑒 = 1

Note that 𝐹 was given to us by direct substitution, which means that in this particular case we have
lim(𝑥 → 𝐺) 𝑓(𝑥) = 𝑓(𝐺) = 𝑓(lim(𝑛 → 0) 𝑔(𝑛))

– – –

EDIT (10/28/21):
The reason this works is because lim 𝑥→0 𝑔(𝑥) = 𝑒 (i.e. the limit exists)
and𝑓(𝑥) is continuous at 𝑥 = 𝑒

According to the theorem for limits of composite functions we then have
lim 𝑥→0 𝑓(𝑔(𝑥)) = 𝑓(lim 𝑥→0 𝑔(𝑥))

Sal explains that theorem here:
• How can e^x be the only function that is the derivative of itself? Doesn't f(x) = 19e^x also satisfy this property?
• When we say that the exponential function is the only derivative of itself we mean that in solving the differential equation f' = f. It's true that 19f = (19f)' but this isn't simplified; I can still pull the 19 out of the derivative and cancel both sides. You are correct in saying that the general solution is Ae^x where A is a real value; however, the "A" part isn't the main focus - the main focus is the exponential, since that's what varies and the constants don't.
• Where can I find the proof of limit as n→infinity (1+1/n)^n =e and limit as n→0 (1+n)^(1/n)=e?
• how/why is (1+1/n)^n equal to (1+n)^(1/n)? Is this just a basic law of exponents
• Think about it like this:

it is completely legal for us to define one variable as some amount of another variable. Therefore, we can say that n=1/u, for example.

Let's say n=1/u

and

(lim n-> inf) e= (1+1/n)^n

Now let's rewrite this in terms of u. The limit will be that u gets very small and approaches 0, because this will cause the fraction 1/u to become very large. For n=1/u: if n approaches infinity, u must approach 0 for both sides to approach infinity.

(lim u-> 0) (1+u)^(1/u) (I simplified 1/(1/u) to just u)

This, therefore, is equivalent to the other definition of e, because all we have done is described the variable in a new way without adding in or taking away anything from the original equation, just looking at it differently.
• At , is it that this is an application of the principle:

lim(x->a)[ f(g(x)) ] = f( lim(x->a)[g(x)] )

?
• Yes, with 𝑓(𝑥) = ln 𝑥 and 𝑔(𝑥) = (1 + 1∕𝑥)^𝑥
we get 𝑓(𝑔(𝑥)) = ln(1 + 1∕𝑥)^𝑥

Because the natural log function is continuous, we have
lim[𝑥 → ∞] 𝑓(𝑔(𝑥)) = 𝑓(lim[𝑥 → ∞] 𝑔(𝑥))
= ln(lim[𝑥 → ∞] (1 + 1∕𝑥)^𝑥)
• Technically, the function x^0-1 is its own derivative.
• Any function of the form a·e^x is its own derivative, and these are the only functions that are their own derivatives. The zero function is just the special case where a=0.
• Hi - i am interested that sal says that e = (1+n)^1/n when I graphed y = (1+x)^1/x the graph converges to 1. What mistake have I made?
• What you may have missed is lim (n->0) for that definition. You are correct that lim (n->∞) (1+x)^1/x = 1, but lim (n->0) (1+x)^1/x = e.
• When/where do we learn that change of variables method?
• At , Sal came up with n . Can the whole proof be shown without this n ? Why did he came up with this idea and not something else ?
• At , how did he change the derivative into a limit?
How is that possible?
What is the formula of changing?
(1 vote)
• That is the definition of derivative as a limit.

The derivative at a point is the slope of the tangent line at that point.

You can verify for yourself that
(𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥))∕𝛥𝑥
is the slope of the line through the points
(𝑥, 𝑓(𝑥)) and (𝑥 + 𝛥𝑥, 𝑓(𝑥 + 𝛥𝑥))

Then, as 𝛥𝑥 → 0 the two points practically become one and the same, and our slope will be that of the tangent line at (𝑥, 𝑓(𝑥)).