Question

What is the derivative of f(x)=3e^(−8x^(2))?

Answer 1

Why am I not good at explaining this?
Derivative of e^u = e^u * du

Answer 2

What will u be in this case?

Answer 3

I'm sorry but what?

Answer 4

Let AccessDenied explain. :)

Answer 5

\( \dfrac{d e^{u(x)}}{dx} = \dfrac{d e^{u(x)}}{du(x)} \times \dfrac{d u(x)}{dx} \)
This is chain rule procedure.
Then you must know the derivative of e^x with respect to x to fill in the details.

Answer 6

Okay I sort of understand that, but I am still a little confuse :s

Answer 7

So, you have this function:
\( \displaystyle y = 3 e^{-8x^2} \)
You can see that this is the exponential function, but it has that weird power of -8x^2 rather than a nice x. So we need to break that up using chain rule, right?

Answer 8

right

Answer 9

So we could call u(x) = -8x^2 That is our entire exponent. Then we use chain rule to take the derivative with respect to u of e^u:
\( \dfrac{d e^{-8x^2}}{dx} = \dfrac{de^u}{dx} = \dfrac{d e^{u}}{du} \times \dfrac{du}{dx} \)
It might look less strange without the u(x) function.

Answer 10

In this way, we just need to know the derivative of the exponential function e^x, and the derivative of whatever was the exponent already.

Answer 11

so the derivative of e^x is e^x, therefore I would take that and apply that to the function u?

Answer 12

That explains the entire part of de^u / du:
\( \dfrac{d e^{u}}{du} = e^{u} \)
The other part is the derivative of u with respect to x, or the derivative of -8x^2 (what we called u)

Answer 13

okay then

Answer 14

\( \dfrac{de^{-8x^2}}{dx} = e^{u} \times \dfrac{d (-8x^2)}{dx} \)
substituting back u =-8x^2
\( = e^{-8x^2} \times \dfrac{d(-8x^2)}{dx} \)

Answer 15

awe okay that makes a lot more sense! thanks!

Answer 16

Glad to help! :)
And generally this is the "formal look" at it, it often becomes mechanical just to know to take the derivative of the inside of the function (-8x^2) and multiply it to the derivative of the parent function.

Also, just a quick note that there was a coefficient of 3 that I intentionally ignored. It still exists and is just tacked on at the end. :)

Answer 17

\( y = \color{red}3 e^{-8x^2} \) that one

Answer 18

on the end you mean it looks like this? \[e^-8x^2 * (d(-8x^2)/dx)*3\]

Answer 19

Yeah. It could also be left at the start with 3e^(-8x^2). The point is, though, that they are just along for the ride and don't do much for the derivative process.
\( \dfrac{dy}{dx} = \dfrac{d}{dx} \left( 3 e^{-8x^2} \right) = 3 \dfrac{d}{dx} \left( e^{-8x^2} \right) \)

Answer 20

awe alrighty then :)

Answer 21

i assume you know how do do the d(-8x^2)/dx part? you don't want to leave it that way at the end. ;p

Answer 22

hmm well no I don't know how to really :s

Answer 23

Just power rule. That is the rule that:
\( \dfrac{d}{dx} x^n = n x^{n-1} \)
There are a lot of rules to know, understandably. So you have a constant to take out again:
\( \dfrac{d}{dx} \left( -8x^2 \right) = -8 \dfrac{d}{dx} x^2 \)
Then d/dx (x^2) is where power rule is used. :)

Answer 24

so, correct me if I'm wrong, based on the power rule the derivative would be d/dx = -16x?

Answer 25

Yes, that is correct.
That would then go in place of the d(-8x^2)/dx in the original problem.

Answer 26

awe alrighty then

Answer 27

\( \dfrac{dy}{dx} = 3e^{-8x^2} (-16x) \)
If you know now how to get to this part, you're doing well.

The biggest thing about derivatives is getting all the rules down. With more practice, this can become a lot easier. Once you understand the rules, you have a lot of access for differentiation. So, good luck on the Maths! :D

Answer 28

haha thanks a lot! this helped a lot!!!