Question

Another calculus problem @zepdrix

Answer 1

Ooo you have all the fun problems tonight huh? XD

Answer 2

lol, yea

Answer 3

Hmm, so you can apply a change of base fancy type thing...
or you can do logarithmic differentiation...
or you can simply recall this derivative identity: \(\large\rm \frac{d}{dx}a^x=a^x(ln a)\)

Which approach? :o

Answer 4

Which way is easiest, lol?

Answer 5

Well obviously identity is the easiest :)
But that doesn't really show where the natural log is coming from.

Answer 6

Okay, lets do logarithmic differentiation them

Answer 7

then*

Answer 8

Boooo

Answer 9

Okay, then, which way do you want to show me? xD

Answer 10

Are you familiar with this type of thing?
\(\large\rm e^{ln x}=x\)
When we take the composition of the exponential operation and the log,
since they're inverses, they simply "undo" one another.
So we can write our x in a fancy way by applying this property in reverse.

Answer 11

If this is too confusing though, we can go to log differentiation :)

Answer 12

yes, I see that.

Answer 13

\[\large\rm \color{orangered}{9^{-x}}=e^{\ln\left(\color{orangered}{9^{-x}}\right)}\]

Answer 14

Ah, that makes sense

Answer 15

Do I need to apply the chain rule?

Answer 16

We'll use our log rule to bring the -x outside,\[\large\rm =e^{-x(\ln 9)}\]From there let's write it in a familiar form:\[\large\rm =e^{(-\ln9)x}\]Which is what we're familiar with in this form right?\[\large\rm e^{ax}\]

Answer 17

(-ln9) is just "a number" attached to the x.
So it wil be that simple chain rule that you're used to.
Just pretend it's the number 2 or something :P

Answer 18

All that fancy business was just a way of rewriting out expression.
We haven't even taken a derivative yet.

Uh oh, the silence :D
I lose ya somewhere in those steps?

Answer 19

So then the answer would be \(\Large -\frac{1}{9}e^{-x(\ln 9)}\)

Answer 20

no, just openstudy lag.

Answer 21

\[\large\rm \frac{d}{dx}e^{\color{orangered}{a}x}=\color{orangered}{a}e^{\color{orangered}{a}x}\]So then,\[\large\rm \frac{d}{dx}e^{\color{orangered}{(-\ln9)}x}=\color{orangered}{(-\ln9)}e^{\color{orangered}{(-\ln9)}x}\]Right? :o
Chain rule gives you this "2" down in front.

Answer 22

HOW DO YOU MAKE IT A DIFFERENT COLOR? IVE BEEN TRYING TO DO THAT FOREVER

Answer 23

But wouldn't we take the derivative of that?

Answer 24

Keep in mind that (-ln9) contains no variable!
It's a `constant`.

He's dressed up in a fancy tuxedo like James Bond, trying to fool you.
His derivative would be 0, not -1/9.

Answer 25

Oh, duh

Answer 26

Our chain rule says to multiply by derivative of (stuff*x)'
which is just =stuff
right? :)

Answer 27

I forgot the x attached to it, so basically it is the derivative of -ln 9 x

Answer 28

Thanks again!

Answer 29

Ok ok but let's make our answer look a little nicer.

Answer 30

k

Answer 31

Remember that big messy exponential thing was our 9^{-x} right?

Answer 32

yes

Answer 33

we can convert it back....?

Answer 34

\[\large\rm \frac{d}{dx}e^{(-\ln9)x}=(-\ln9)\color{orangered}{e^{(-\ln9)x}}\qquad=\qquad (-\ln9)\color{orangered}{9^{-x}}\]Ya that would be a good idea :)

Answer 35

It's good to see this process at least once.
Maybe from now on you just remember the identity.\[\large\rm \frac{d}{dx}9^{-x}=9^{-x}(\ln 9)(-x)'\]

Answer 36

You might ask yourself...
"wait wait wait... why doesn't e^x follow this rule?"

Any ideas? :)

Answer 37

So, wait, is this showing me the reason behind the idenity/formula you showed me at the beginning?

Answer 38

Mmm, yes I suppose so XD

Answer 39

And e^x doesn't follow this rule because the derivative of it is just itself...?

Answer 40

Logarithmic differentiation would have done the same,
and taken mehh about the same amount of time.

Answer 41

The answer is, e^x actually DOES follow this rule,
it's just that (ln e) =1 so we don't bother writing it.

Answer 42

\[\large\rm \frac{d}{dx}e^x=e^x(\ln e)\]

Answer 43

But it's good to understand that all of our exponentials follow this rule :)
ya?

Answer 44

Ah, I see.

Answer 45

Because like you said, they're opposites right?

Answer 46

I guess it's more based on how we define the log operation :)\[\large\rm \log_{10}(\color{royalblue}{10})=x\]This x represents the power that our base needs to be raised to in order to end up with this blue 10.

So with natural log,\[\large\rm \ln_e(\color{royalblue}{e})=x\]This x represents the power our base (e) needs to e raised to in order to end up with this blue e.

Answer 47

e to the `first power` is what gives us e, yes?
So this x value is 1.

Answer 48

needs to be* raised to
ahh that was an unfortunate typo -_-

Answer 49

yes.

Answer 50

so ya, natural log of e is 1 based on how log works :)
blah blah blah

Answer 51

But yes, you could think of it in terms of the "opposites" thing,
if you realize that e = e^1\[\large\rm \ln(e^{\color{orangered}{1}})=\color{orangered}{1}\]

Answer 52

lol, yep. Thanks for the explanation. :)

Answer 53

ok ok ill stop rambling :D
calc is so much fun though!!

Answer 54

lol, you're hilarous. I think you and solomonzelman are the only people on here that really love math that much. I think it's wonderful how much you know

Answer 55

kainui and dan are pretty enthusiastic as well XD
they're way too smart for my liking though lolol

Answer 56

lol, I haven't interacted with them much, so I can't say.