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.

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