Why is the derivative of e^x equal to e^x?

In this video, we will learn the derivatives of exponential functions and we will see how we can define the number e. Calculus 1, AP calculus, calculus A-level lesson.

0:00 We will talk about why the derivative of e to the x is e to the x
0:30 Derivative of 2^x by the definition of derivative
3:43 Defining the number e
8:07 Differentiate b^x
9:52 Check out Brilliant
10:51 Bonus: derivative of ln(x)
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[teekak]

Fun fact, based on this definition you can actually rearrange the equation to arrive at the definition of e based on compound interest:
lim(h-->0)((e^h - 1)/h) = 1
Let h_0 be a small number such that
(e^h_0 - 1)/h_0 = 1
e^h_0 - 1 = h_0
e^h_0 = 1 + h_0
e = (1 + h_0)^(1/h_0)
Convert back to limit
e = lim(h-->0)(1+h)^(1/h)
i.e. compound interest recursion formula with infinitesimally small interest rate for an infinitely large number of compounding periods.

[ignantxxxninja]

As a math major, I really enjoy how you explain these things. A lot of these videos are not trivial at all

[AliKhanMaths]

It's really important to understand that it's not that the derivate of e^x is just e^x, but that's WHY the number e exists. Videos like yours really inspire me to share my own videos as well!

[gordonglenn]

As a veteran calculus teacher, I enjoy learning something from almost every BPRP video. Thank you!

[vari]

I love how this shows the process of discovering the derivatives and defining things along the way; it makes math much more interesting when you see how these things might’ve been originally figured out.

[ЮрійЯрош-гь]

But before defining e in such a way, you need to prove that there exists a unique number with the described property and I can't say that that's quite easier then proving that the derivative of e^x is e^x using some other definition of e.

[garyhuntress]

For a few minutes I thought you were intentionally breaking up the white board into golden ratio rectangles!

[michapodlaszuk]

I'm preparing for math exams in may and this is the kind of content I need right now, I'll pray for more blackpenredpen content suggested for me by the youtube algorithm 🙏

[staswisniewski]

3B1B made also video about this in his Calculus Series.

[pacolibre]

You can actually evaluate the limit of (2^h-1)/h by using the variable substitution u=2^h-1, turning it into the limit as u->0 u/log2(u+1) = 1/((1/u)log2(u+1)) =1/log2((1+u)^(1/u)).

Change of variables to t=1/u gives lim t->inf 1/log2((1+1/t)^t). This includes the limit definition of e which can be shown to exist by the monotone sequence theorem.

Then you have 1/log2(e) =log2(2)/log2(e) = ln(2) by the change of base formula.

I found out about this from Khan Academy.

[Morbius_Official]

Perfectly executed! I love how you managed to connect everything so wonderfully and I could see the beauty in it.

[GPLB]

Thank you! This video helped me realize the relationship to define ln(a) as the limit of (a^h-1)/h as h goes to 0.

[bhavydugar]

Finally after 2 short videos one long video
Really innovative and cool 😎
I really thought you were gonna say d/dx(2^x)=e^x and 0.2 is even

[Timeflow_X]

for the exponential function b^x, the derivative by first principles becomes (b^x) lim h→0 ((b^h) - 1)/h = (b^x) lim h→0 ((b^h) - (b^0))/h, which is the instantaneous gradient at x = 0, and b = e is the only case where the gradient at x = 0 is 1

idk, I think it's a fun visualization of what that weird limit is actually representing

[altiarissuralis]

Funny enough if you arrange the limit (e^h-1)/h =1 as h—>0, you get e=(1+h)^(1/h) as h—>0.

e=(1+h)^(1/h) as h—>0 is a generally used interest formula, that is continuous growth.

[krischan]

I think when I learned it at school (in the math power lessons at grammar school with a teacher who had a math PhD), we started by pointing out that f(x)=1/x must have an antiderivative F (I'm avoiding to use the terms ln, e etc.). This cannot be calculated by using the (x^n)' = n*x^(n-1) formula because it would involve a division by 0, respectively it doesn't work for n=0. We chose to calculate the one with the constant leading to F(1)=0 and then made deductions about its inverse function G. We showed that G'=G, that G is of the form G(x) = Z^x by proving that for every x1 and x2, G(x1+x2)= G(x1)*G(x2). Finally we calculated G(1) = Z^1 = Z (which ie and voila, we were done.

[noordinbashir]

Finally someone who explained it properly. Really helpful video.

[shehnazsalahuddin]

Oh! I wanted to know about this for a long time. Thank you so much bprp for this tutorial!

[gibbogle]

Great stuff! Maths students today are so lucky to have the internet with great teachers like this man.

[SimplisticVR]

You could also find e^x sum definition and d/dx the sum aka taking the derivative of each fraction