lordofjudgement reblogged your post and added:

Is 18x^2(e^4x^3), in truth 18x^2(e^6x^3) and you made a typo?

Because then 18x^2(e^6x^3) is the correct answer. Not sure where you got 4 XD

Derivative of  e^6x^3 is  e^6x^3 multiplied of the derivative of x^3 is 6*d(x^3)/dx = 6*3X^2 = 18X^2.     (/= divided).

Which gets the answer:   18x^2(e^6x^3

Yeah oops I made a typo lool

Umm I didn’t really follow that. Does that follow from a rule I’m forgetting? Why isn’t the derivative (e^kx)(k)?

baddaytobuywine replied to your post “moar maths  So this notation will be confusing. But basically it’s e…”

There is no need for the chain rule, or special rules for second exponents. It is a exponential deriv. So it is in a^x you go a^x*ln a*deriv of x which solves that to the answer. With chai you have the deriv of e which is zero so the whole thing would be0

No, ln isn’t in the answer. The correct answer is 18x^2(e^4x^3).

dustyflames reblogged your post and added:

Nah, your answer is wrong. 

When you differentiate e^(Watev), you get d(Watev)*e^(Watev).

So you’ll get : 6*(3)*x^2*e(6*x^3)

The chain rule doesn’t apply.

Okay i wasn’t that far off tho so

I thought differentiating e^kx was e^kx times k, and differentiating e^x is just e^x. So that can’t be true for all e^(whatevers), because those are two different instances where it’s not the derivative of whatever times e to the whatever. 

So I’m confused. I’m pretty sure the chain rule is supposed to be at play because it’s a chapter on it.

I am probably forgetting some rule I need to be reminded of.

Proof of Euler's Identity

I was wondering if anyone could help me with this. I’m trying to come up with a proof of Euler’s Identity using derivatives rather than the Taylor polynomials, and this is what I have so far, but I don’t think it’s quite rigorous:

e^ix is a complex number, so it has a real and imaginary part, and both can be written as some real-valued function of x:

e^ix = f(x) + ig(x)

Taking the derivative of both sides gives:

i(e^ix) = f’(x) + ig’(x)

Multiplying both sides by -i gives:

e^ix = -if’(x) + g’(x) = f(x) + ig(x)

Separating real and imaginary parts gives:

-if’(x) = ig(x)
g’(x) = f(x)

The (I want to say only, but I’m not certain.) nontrivial functions that satisfy both f’(x) = -g(x) and g’(x) = f(x) are f(x) = cos(x) and g(x) = sin(x). So,

e^ix = cos(x) + isin(x)

Is this proof conclusive, or am I making an unjustified leap somewhere since I already know Euler’s Identity?

3

Gabriel’s Horn has an infinite surface area and a finite volume, which is appropriately representative of its symbolic connection between the two.

i did calculus for 16 hours today. i feel better about it but there’s still so much i need to improve. i’m trying to come to terms with the fact that i’m not naturally good at math so i need to work like 5x as hard as other people to do well, but in the end it will all be worth it.

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