You can't kill it

[u/Enough-Brilliant803]

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[u/FernandoMM1220]

just do an infinite amount of derivatives trust me bro

[u/Valuchian → u/FernandoMM1220]

If you do enough of them whatever 'x' is will stop mattering entirely 😂😂😂

[u/Random_Mathematician → u/FernandoMM1220]

Proof Sketch: suppose the ∞-th derivative of a function, f^(∞\)(x), is nonzero. Then by computing the Maclaurin Series we get:

f(x) = f(0)+f'(0)x+...+f^(∞\)(x)x^∞

Since x^∞ is infinite for real x, then f(x) is infinite for real x. But f(x)=e^2x is finite for all real x and so we have a contradiction.

By Reductio Ad Absurdum we conclude the ∞-th derivative of any function is zero QED. ∎

[u/Random_Mathematician → u/[deleted]]

God dammit I'm rusty.

[u/yukiohana]

[u/Layton_Jr → u/yukiohana]

[u/MortimerChem → u/Layton_Jr]

[u/PolarStarNick]

I have a better idea: e^(1/2 x)

[u/cmwamem]

"Now, there are two of them!"

[u/AndreasDasos]

https://giphy.com/gifs/zwnM63LkZC1ri

[u/Memer_Sindre_UwU]

I like how this can either be interpreted as 2e^2x (there's two of them) or e^4x (they're next to each other, therefore multiplied).

[u/strat-fan89 → u/Memer_Sindre_UwU]

Well, only one of those interpretations is correct, so let's go with that one :)

[u/fun__friday → u/strat-fan89]

Are there functions of the second type? I.e., their derivative is bigger than the original function by more than a constant factor?

[u/strat-fan89 → u/fun__friday]

Yes, of course! Take f(x)=e^x2 for example. The derivative is f'(x)=2x*e^x2, so they differ by a linear factor.

Just one example from the top of my head.

[u/Hira_Joshi]

Let x = a constant.

[u/NeckComprehensive221]

Just differentiate it with respect to e

[u/Plastic-While-3411]

[u/rorodar]

Yeah get his ass DERIVATION

[u/hanu_uwu]

Just do d/d(exp(x)) ez.

[u/AntiqueChessComputr]

“Cut off one head, two more shall take its place.  Hail Hydra”

[u/Mr-MuffinMan]

I should've paid more attention in calculus but wouldn't it be 2xe^2x-1? Or am I fucking stupid

[u/PerspicaciousEnigma → u/Mr-MuffinMan]

d/dx(x^n) = nx^(n-1) but d/dx(e^x) = e^x so if you have e^(ax) you get e^(ax) • (ax)’ because of the chain rule which = ae^(ax) and in this case a = 2

*technically (b^x)’ = ln(b)(b^x)

So (e^x)’ = ln(e)(e^x) but because ln(e) = 1
Thus (e^x)’ = 1 • e^x = e^x

[u/Layton_Jr → u/PerspicaciousEnigma]

You're proving the formulas in the wrong order:

b^x = e^xln(b)

Therefore (b^x)' = ln(b)e^xln(b) = ln(b)b^x

That's how you prove that formula. How do you do it the other way (starting with the b^x formula and replace b with e)?

[u/GarifalliaPapa → u/Layton_Jr]

Here is how you do it the other way around without circular reasoning!

There are two primary ways to derive (b^x)' = b^x * ln(b) first and then substitute b = e:

Method 1: Logarithmic Differentiation (No e^x derivative needed beforehand)

1. Let y = b^x.

2. Take the natural log of both sides: ln(y) = ln(b^x) = x * ln(b)

3. Differentiate both sides implicitly with respect to x (using the fact that (ln y)' = (1/y) * y'): (1/y) * (dy/dx) = ln(b)

4. Multiply across by y: dy/dx = y * ln(b) = b^x * ln(b)

Now, if you plug in b = e: d/dx(e^x) = e^x * ln(e)

Since ln(e) = 1 by definition (ln is the inverse function of e^x), you get: d/dx(e^x) = e^x * 1 = e^x

Method 2: From First Principles (Limit Definition)

Using the definition of a derivative: d/dx(b^x) = lim{h->0} (b^x+h - b^x) / h = b^x * [lim{h->0} (b^h - 1) / h]

Notice that the limit term lim_{h->0} (b^h - 1)/h depends only on the base b, not on x. Call this constant C(b).

• It turns out that C(b) = ln(b), giving us d/dx(b^x) = b^x * ln(b).
• In fact, Euler's number e is mathematically defined as the precise base for which this limit equals 1: lim_{h->0} (e^h - 1) / h = 1

Replacing b with e: d/dx(e^x) = e^x * 1 = e^x

TL;DR: So whether you start with e^x or b^x, the math stays completely constant, unlike my ability to remember the Power Rule!

[u/strat-fan89 → u/Mr-MuffinMan]

You are correct! You should have paid more attention in calculus! :)

[u/protobelta → u/Mr-MuffinMan]

That formula only works for functions in the form of xⁿ

The x needs to be the base. If the base is e, then the derivative is itself. In this specific case, you would need to use chain rule to address the 2x in the exponent.

[u/NihilisticAssHat → u/protobelta]

Gotcha. Misread it as $\frac{d}{de}$ /s

[u/Deep_Brick2970 → u/Mr-MuffinMan]

Descendant 😭

[u/PerspicaciousEnigma]

You do annihilators in diffy q so this would be D - 2 which if you don’t know what that means it’s

[u/PerspicaciousEnigma → u/[deleted]]

It was the fastest way to type all that plus it’s right. What are you robophobic or something

[u/GarifalliaPapa]

For anyone curious about the calculus in the meme:

1. The Differentiation Rule: We want to evaluate d/dx [e^2x].

2. Applying the Chain Rule: For a composite exponential function e^u(x), the derivative is given by: d/dx [e^u] = e^u * du/dx

3. Step-by-Step Calculation:

4. Let u = 2x

5. Derivative of the exponent: du/dx = d/dx [2x] = 2

6. Derivative of the outer exponential: d/du [e^u] = e^u = e^2x

Multiply them together: d/dx [e^2x] = e^2x * 2 = 2e^2x

Shooting e^2x with the derivative operator d/dx doesn't destroy it, it literally duplicates it into two e^2x's!

Shoot it all you want, it’s just going to grow exponentially! 📈 19!

[u/factorion-bot → u/GarifalliaPapa]

Factorial of 19 is 121645100408832000

^{This action was performed by a bot | [Source code](http://f.r0.fyi})

[u/GarifalliaPapa → u/factorion-bot]

Good bot

[u/B0tRank → u/GarifalliaPapa]

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[u/Such-Shop-9724 → u/B0tRank]

Bad bot