anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
You already have all the tools you need for differentiation. Do you have a specific example?
zepdrix
  • 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
  • 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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zepdrix
  • zepdrix
oh lol :)
zepdrix
  • 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
  • anonymous
lol
anonymous
  • anonymous
lol @zepdrix has explained it better than i would
zepdrix
  • 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
  • anonymous
\[e^{2x}\div e^{4x ^{2}} + \cos(e ^{7x})\]
anonymous
  • anonymous
it approaches infinity?
anonymous
  • anonymous
that was me answering the post before i posted that big equation @zepdrix
anonymous
  • anonymous
i think i answered wrong lol
zepdrix
  • 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
  • zepdrix
What do we need to do with the big messy problem? Take it's derivative?
anonymous
  • 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
  • anonymous
And differentiate it
anonymous
  • anonymous
taking the derivative and differentiating are the same thing right?
zepdrix
  • zepdrix
Yes, just fancy words ^^
anonymous
  • anonymous
ok that's what i thought lol
anonymous
  • 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
  • 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
  • zepdrix
Ok ok big question time c:
anonymous
  • anonymous
ok that makes sense = ) ty! and yes now to the big one lol
zepdrix
  • zepdrix
handy*
anonymous
  • anonymous
hold on lol cos is in the denominator with e to the 4x^2
zepdrix
  • zepdrix
Oh i see ^^
zepdrix
  • 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
  • zepdrix
\[\large \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\]
anonymous
  • anonymous
right that's what they did with it = ) that's where I got confused, not with the actual rule but u'
anonymous
  • anonymous
and v'
zepdrix
  • 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
  • zepdrix
Remember how to do the chain rule? c:
anonymous
  • anonymous
yes
zepdrix
  • zepdrix
\[\huge \left(e^{2x}\right)'=e^{2x}(2)\]Any confusion on that one? c:
anonymous
  • anonymous
oh wow i think i just had a mental click lmfao
zepdrix
  • zepdrix
lol XD those are always fun.
anonymous
  • anonymous
hold on let me look at the problem and answer one more time lol
anonymous
  • 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
  • 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
  • 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
  • anonymous
oh and then one more after that lol but i think it may be pretty easy haha
zepdrix
  • 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
  • zepdrix
Since the base is LARGER than 1, and our exponent is positive, the value will explode.
zepdrix
  • zepdrix
What if you had,\[\large \lim_{x \rightarrow \infty} (.999)^x\]
anonymous
  • anonymous
so really big x values = really big y values
zepdrix
  • zepdrix
yah c: As x approaches infinity, y is approaching infinity. In that first case.
anonymous
  • anonymous
I used a calculator but the bigger x gets the closer y goes to 0 for the second part
zepdrix
  • 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
  • zepdrix
Yesss good good c:
anonymous
  • anonymous
correct = )
anonymous
  • anonymous
so in that case it would go to -infinity
zepdrix
  • zepdrix
That's a good strategy with limits. If you're ever confused, just plug numbers in to see what happens.
zepdrix
  • zepdrix
Noooo :D An exponential can't go in the negative! :O
anonymous
  • anonymous
or wait it would go to 0 not -infinity right bc yeah what you just said haha
anonymous
  • 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
  • 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
  • zepdrix
Ok simmer down :O Let's look at the first one a sec.
anonymous
  • anonymous
lol
zepdrix
  • 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
  • anonymous
yes
zepdrix
  • 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
  • zepdrix
\[\large 4=a^2 \qquad \rightarrow \qquad a=2\]
anonymous
  • anonymous
ok that makes sense = ) is that a property or a rule?
zepdrix
  • zepdrix
It's this really fancy thing called "division" :3
zepdrix
  • zepdrix
lolol
anonymous
  • anonymous
or what i mean to say is how did you know to divide the points' equations hahahaha
zepdrix
  • 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
  • zepdrix
We could have also done substition to get the same result though :)
anonymous
  • anonymous
like solve for C in one of the problems and then plug it into the other equation?
zepdrix
  • zepdrix
Yah :)
anonymous
  • anonymous
ok good i think i got it = ) thank you so much for all your help! = D
zepdrix
  • zepdrix
yay team \c:/
anonymous
  • anonymous
lol

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