calculus

>>8450373
[eqn]\int e^x[/eqn] hehe

>>8450373
what is the derivative of e^x??

>>8450373
because [math]{d\over dx}e^u=e^u[/math]

duh

>>8450375
69

It is the function whose derivative is itself
y'=y

>>8450373
Nobody knows, it's a mystery.

Because it just is. Prove it to yourself using Riemann sums.

taylor expand and derive u will see

>>8450373
the integral is the accumulation of the function; hence why you can use the definite integral to find the area under a curve

ration it like this; every point on the graph of e^x represents BOTH the slope of the tangent line of the function as well as the accumulation (sum of infinitesimals) up to that x value. hence why it is its own derivative and integral

>>8450373
take its derivative

if that doesn't make sense use the limit definition of a derivative

if you don't understand why the integral is the antiderivative, do a riemann sum

integrate the taylor series and it'll make more sense

>>8450373
It's its own grandfather m80

draw a graph of e^x and you'll quickly find out you need e^x of lead to colour under the graph.

File: 3472.jpg (64KB, 630x800px) Image search: [Google]

64KB, 630x800px

>>8450379
>dx
>u

>>8450373
Math is more pointless than videogames.

you dont need to prove it faggot. youre in calc 2

>>8450373
How are you defining [math] \exp x [/math]?
Using this?
[eqn] \ln x = \int^1_0 \frac{1}{x} [/eqn]
Or this?
[eqn] \exp x = \sum^\infty_{i=0} \frac{x^i}{x!} [/eqn]

*pardon any mistakes. I just woke up and can't be arsed to look up my taylor series.

>>8450373
Remember that e is a constant number. exp(1) is ~2.718

Also, you're integrating in respect to x. e is not a variable.

File: 1417979938773.jpg (68KB, 529x595px) Image search: [Google]

68KB, 529x595px

>>8450379
>{d\over dx}
>this actually works
fugg

File: tegaki.png (13KB, 400x400px) Image search: [Google]

13KB, 400x400px

>>8450373

>>8450373

Well done op, nice bait.

>>8450734
I hope you're enjoying that electronic device that's doing billions of pointless things per second.

File: a88.png (48KB, 500x514px) Image search: [Google]

48KB, 500x514px

>>8450774
You spent more time TeXing than thinking, didn't you?

>>8450373
Idk how to use the math plugin, I'm dumb.
Axioms:
d/dx ln(x) = 1/x
d/dx x=1

d/dx ln(e^x) = d/dx x=1
d/dx ln(e^x) = d/du ln(u) d/dx e^x //u = e^x
= 1/u d/dx e^x
= 1/e^x d/dx e^x
= 1
Q.E.D

>>8450373
Could copy my last year notes with comments, but too lazy brah

Isn't it basically defined that way?

>>8451980
Yes and no. It more so happens that the L(x) e^x, dy/dx e^x and the integral of e^x dx are all e^x, as such it is incredibly useful for hyperbolic series.

because e is the transcendental connective tissue between functions of different powers.

You stupid fat fuck

>>8450774
cease and desist

>>8450373
it doesnt.

[math]\int e^x dx = e^x + c[/math]

>>8453478
What OP wrote is true because what you wrote is true. OP isn't wrong.

>>8453485
OP gave a particular solution.
i personally dont think that is sufficient.

>>8450373
> the integral of a function
> no precision of variables type
Assuming it's the real defined and valued function
> no bounds
Stopped thinking about it. Reformulate your question.

>>8451977
>intelligent af
>too lazy bruh
literally every /sci/ user

>>8450373
e^x = 1+x+x^2/2!+x^3/3!+x^4/4! ...

int e^x dx = 1+x+x^2/2!+x^3/3!+x^4/4! ... + c' || c' = c-1

int e^x dx = e^x + c'

Did not bother to format in latex because it is this easy to show. The constant is c' for absolute dummies who can't see "where the 1 goes".

ITT: op trolls people into writing 'sex' a bunch of times

>>8450379
>>8450732
[math]\int_{X}^{} a_uf_udu[/math]

>>8450373
>>8450375
Please write the whole sentence anon:
[math] \int e^{x} = f(u)^{n} [/math]
.
.
Proof on request.

>>8456565
Prove it.

>>8456666
>>8456565

Let [math]f(a)=(e^x)^{1/n}[/math].
.
[math]\lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{f(t)} \right)=\frac{d}{dx}f(u) \Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{\infty} \right) [/math]
[math] = \frac{d}{dx}f(u) \Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - 0 \right) [/math]
[math]= \frac{d}{dx}f(u).[/math]
.
That yields:
.
[math](e^x)^{1/n} = \frac{d}{dx}f(u) \Leftrightarrow e^x[/math]
[math] = \frac{d}{dx}f(u)^n \Rightarrow \int e^x [/math]
[math] = \int \frac{d}{dx}f(u)^n.[/math]
.
Let's conclude:
.
[math]\int e^x = f(u)^n[/math].

>>8456801
>>8456666
>>8456565
Better display.

Let [math]f(a)=(e^x)^{1/n}[/math].
.
[math]\lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{f(t)} \right)=\frac{d}{dx}f(u) [/math]
[math]\Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - \frac{1}{\infty} \right) = \frac{d}{dx}f(u) [/math]
[math]\Leftrightarrow \lim_{t \rightarrow \infty} \left( f(a) - 0 \right) = \frac{d}{dx}f(u).[/math]
.
That yields:
.
[math](e^x)^{1/n} = \frac{d}{dx}f(u) \Leftrightarrow e^x = \frac{d}{dx}f(u)^n [/math]
[math]\Rightarrow \int e^x = \int \frac{d}{dx}f(u)^n.[/math]
.
Let's conclude:
.
[math]\int e^x = f(u)^n[/math].

>>8456801
what a complete bullshit proof.

File: 4gGrEC9.gif (2MB, 300x150px) Image search: [Google]

2MB, 300x150px

>>8456836
Did you get this is a meme?

Have you even understood there is a sentence involved in the conclusion of this "proof"?

Because you seem to be such a retard, I am giving myself the right to call you a faggot.

>>8457309
yes ofcourse i get what you are saying retard.

sex is fun hahaha what a nice meme.

still you could have put in effort to make an actual proof. not just that bogus you wrote down.
kys

>>8457331
dumb cs fag cant read math..
heres something in your language.. haha
you == fail && me.KEK() ? die whore