Can anyone explain how to differentiate exponential functions? I have the Calc 1 basic knowledge so I know the main differentiation rules (ie. 4x^2=8x) and I also know the addition, subtraction, multiplication, and division rules as well

- anonymous

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- schrodinger

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- anonymous

You already have all the tools you need for differentiation. Do you have a specific example?

- zepdrix

\[\large y=2^x\]Hmm so the variable is in the exponent position.
That poses a problem.
We can't apply the Power Rule that you may have learned at this point.
We have to introduce logarithms in order to get the variable OUT of the exponent position.

- anonymous

I have e^5 and it says the answer is 0 but I don't understand why that is. and there are a couple others I don't get. Bc my book says that the exponential function is itself. Why wouldn't it be just f'(x)= e^5?

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## More answers

- zepdrix

oh lol :)

- zepdrix

They're pulling a trick on you. See how e^5 contains no variable? It's just a fancy fancy looking CONSTANT, all dressed up in a suit.

- anonymous

lol

- anonymous

lol @zepdrix has explained it better than i would

- zepdrix

\[\large f(x)=e^5 \qquad \qquad \rightarrow \qquad \qquad f(x)=148.4\]See how the derivative is going to turn out? :)
What happens when you take the derivative of a constant?

- anonymous

\[e^{2x}\div e^{4x ^{2}} + \cos(e ^{7x})\]

- anonymous

it approaches infinity?

- anonymous

that was me answering the post before i posted that big equation @zepdrix

- anonymous

i think i answered wrong lol

- zepdrix

Nooo silly! :O
That's one of those main derivative rules that you want to know at this point! :)
A constant will give you 0 when you differentiate it.
A derivative measures change.
A constant is something that stays constant, doesn't change.
So how much does `something that doesn't change` ... change?
zerooooOooOO!

- zepdrix

What do we need to do with the big messy problem? Take it's derivative?

- anonymous

That's right sorry, see the problem i have is answering questions that are put in to words haha but a lot of people struggle with that, exactly why i hate optimization problems = ( but yes derivative of a constant is 0. So why wouldn't it be 1 then? Isn't anything to the 0 power 1?

- anonymous

And differentiate it

- anonymous

taking the derivative and differentiating are the same thing right?

- zepdrix

Yes, just fancy words ^^

- anonymous

ok that's what i thought lol

- anonymous

So I have the answer for the big equation from my neighbors notes but it doesn't make sense to me so that's why I want to ask about it = )

- zepdrix

We can think about it like this, maybe this will help, maybe not :)\[\large f(x)=e^5\]See how there is no X value in the equation? Let's introduce an X.\[\large f(x)=(e^5)x^0\]
\(x^0\) is just \(1\) right? So we're allowed to do that. We're just multiplying it by 1.
Taking the derivative (applying the power rule to the x term) gives us,\[\large f'(x)=0(e^5)x^{0-1}\]We have a 0 coming down be multiplied by everything, that's going to turn the whole thing to 0.
Maybe you're just getting confused by the e term.
It contains no X's. We won't be trying to take it's derivative by applying the \(e^x\) rule.

- zepdrix

Ok ok big question time c:

- anonymous

ok that makes sense = ) ty! and yes now to the big one lol

- zepdrix

handy*

- anonymous

hold on lol cos is in the denominator with e to the 4x^2

- zepdrix

Oh i see ^^

- zepdrix

\[\huge \frac{e^{2x}}{e^{4x^2}+\cos(e^{7x})}\]Ah ok so it is messy :)
We'll need to apply the quotient rule.

- zepdrix

\[\large \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\]

- anonymous

right that's what they did with it = ) that's where I got confused, not with the actual rule but u'

- anonymous

and v'

- zepdrix

\[\huge \left(e^{2x}\right)'=e^{2x}(2x)'\]So we'll apply the e^x rule, but since we have more than just x in the exponent, we have to apply the chain rule.
The prime on the brackets is to show that we need to take it's derivative still.

- zepdrix

Remember how to do the chain rule? c:

- anonymous

yes

- zepdrix

\[\huge \left(e^{2x}\right)'=e^{2x}(2)\]Any confusion on that one? c:

- anonymous

oh wow i think i just had a mental click lmfao

- zepdrix

lol XD those are always fun.

- anonymous

hold on let me look at the problem and answer one more time lol

- anonymous

omg I'm so dumb lol Idk if its bc how you explained it or if I was just looking at it blindly lol omg you were such a great help either way lol

- zepdrix

Chain rule can be a little confusing at first. You're essentially making a 'copy' of the inner function and then taking its derivative, which can seem a little strange c:

- anonymous

but I do have another one lol
it says \[\lim_{x \rightarrow \infty} (1.001)^{x}\]
and it says the answer is infinity, why is that?

- anonymous

oh and then one more after that lol but i think it may be pretty easy haha

- zepdrix

Just plug in a really big value for x and you'll see where it's heading! :)
If x=99999\[\large 1.001^{99999}=2.55 \times 10^{43}\]

- zepdrix

Since the base is LARGER than 1, and our exponent is positive, the value will explode.

- zepdrix

What if you had,\[\large \lim_{x \rightarrow \infty} (.999)^x\]

- anonymous

so really big x values = really big y values

- zepdrix

yah c:
As x approaches infinity, y is approaching infinity. In that first case.

- anonymous

I used a calculator but the bigger x gets the closer y goes to 0 for the second part

- zepdrix

In this second case, see how the BASE is less than 1?
When we raise it to larger and larger powers it will get smaller and smaller and smaller.
Imagine multiplying fractions. They get smaller! c:
\[\large \dfrac{1}{2}\cdot \dfrac{1}{2} \quad = \quad \dfrac{1}{4}\]

- zepdrix

Yesss good good c:

- anonymous

correct = )

- anonymous

so in that case it would go to -infinity

- zepdrix

That's a good strategy with limits. If you're ever confused, just plug numbers in to see what happens.

- zepdrix

Noooo :D An exponential can't go in the negative! :O

- anonymous

or wait it would go to 0 not -infinity right bc yeah what you just said haha

- anonymous

ok and so for the last one, well at least for now haha
It says find the exponential function f(x)=Ca^{x} whose graph is given.
It then has two problems... it shows you a graph from the exponent family and gives you two points on the graph, the first problem gives you the points: (1,6) and (3,24)
I set the problem up as y=Ca^{x} and plugged 6 (the first y-value) in for y and then plugged 1 into the x. It says the answer is 3*2^{x}. So then I looked at the second point. I did the same thing as I did with the first pair but I couldn't see how you could figure it out without doing trial and error.
The second problem gave the two points (2,2/9) and a y-intersect at 2. So I tried the same approach with the first pair and got 2*(1/3)^{x} as the answer but I'm not sure if that's correct

- anonymous

I feel the second answer I arrived at might be correct because a fraction with an exponent has a graph that is decreasing which the graph is decreasing in the picture of the second problem's graph @zepdrix

- zepdrix

Ok simmer down :O Let's look at the first one a sec.

- anonymous

lol

- zepdrix

\[\large f(x)=Ca^x\]\[\large f(1)=6 \qquad \rightarrow \qquad 6=Ca^1\]\[f(3)=24 \qquad \rightarrow \qquad 24=Ca^{3}\]So this is where you got stuck on the first one?

- anonymous

yes

- zepdrix

We have to perform a little sneaky trick at this point!
We'll DIVIDE our f values.\[\large \frac{f(3)}{f(1)} \qquad \rightarrow\qquad \frac{24}{6}=\frac{Ca^3}{Ca^1}\]We get a nice cancellation with the C's allowing us to solve for a.\[\large \frac{24}{6}=\frac{\cancel{C}a^3}{\cancel{C}a^1}\]

- zepdrix

\[\large 4=a^2 \qquad \rightarrow \qquad a=2\]

- anonymous

ok that makes sense = ) is that a property or a rule?

- zepdrix

It's this really fancy thing called "division" :3

- zepdrix

lolol

- anonymous

or what i mean to say is how did you know to divide the points' equations hahahaha

- zepdrix

Well here is something to think about.
When you get to this point,\[\large 6=Ca^1\]\[\large 24=Ca^3\]
You have a SYSTEM of equations.
To be more specific, you have 2 equations and 2 unknowns.
Since the number of unknowns does not exceed the number of equations given, we can find solutions for each unknown.
Remember back to algebra, doing Substitution and Elimination?
This is just another silly trick to add to your list.
I'm not really sure what it's called though XD

- zepdrix

We could have also done substition to get the same result though :)

- anonymous

like solve for C in one of the problems and then plug it into the other equation?

- zepdrix

Yah :)

- anonymous

ok good i think i got it = ) thank you so much for all your help! = D

- zepdrix

yay team \c:/

- anonymous

lol

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