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anonymous
 3 years ago
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
 3 years ago
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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You already have all the tools you need for differentiation. Do you have a specific example?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large y=2^x\]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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1They'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
 3 years ago
Best ResponseYou've already chosen the best response.0lol @zepdrix has explained it better than i would

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large f(x)=e^5 \qquad \qquad \rightarrow \qquad \qquad f(x)=148.4\]See how the derivative is going to turn out? :) What happens when you take the derivative of a constant?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[e^{2x}\div e^{4x ^{2}} + \cos(e ^{7x})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it approaches infinity?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that was me answering the post before i posted that big equation @zepdrix

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think i answered wrong lol

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Nooo 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
 3 years ago
Best ResponseYou've already chosen the best response.1What do we need to do with the big messy problem? Take it's derivative?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That'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
 3 years ago
Best ResponseYou've already chosen the best response.0taking the derivative and differentiating are the same thing right?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, just fancy words ^^

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok that's what i thought lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.1We 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,\[\large f'(x)=0(e^5)x^{01}\]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
 3 years ago
Best ResponseYou've already chosen the best response.1Ok ok big question time c:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok that makes sense = ) ty! and yes now to the big one lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold on lol cos is in the denominator with e to the 4x^2

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \frac{e^{2x}}{e^{4x^2}+\cos(e^{7x})}\]Ah ok so it is messy :) We'll need to apply the quotient rule.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large \left(\frac{u}{v}\right)'=\frac{u'vuv'}{v^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right that's what they did with it = ) that's where I got confused, not with the actual rule but u'

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \left(e^{2x}\right)'=e^{2x}(2x)'\]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
 3 years ago
Best ResponseYou've already chosen the best response.1Remember how to do the chain rule? c:

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \left(e^{2x}\right)'=e^{2x}(2)\]Any confusion on that one? c:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh wow i think i just had a mental click lmfao

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1lol XD those are always fun.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold on let me look at the problem and answer one more time lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0omg 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
 3 years ago
Best ResponseYou've already chosen the best response.1Chain 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
 3 years ago
Best ResponseYou've already chosen the best response.0but 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh and then one more after that lol but i think it may be pretty easy haha

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Just 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
 3 years ago
Best ResponseYou've already chosen the best response.1Since the base is LARGER than 1, and our exponent is positive, the value will explode.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1What if you had,\[\large \lim_{x \rightarrow \infty} (.999)^x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so really big x values = really big y values

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1yah c: As x approaches infinity, y is approaching infinity. In that first case.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I used a calculator but the bigger x gets the closer y goes to 0 for the second part

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1In 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: \[\large \dfrac{1}{2}\cdot \dfrac{1}{2} \quad = \quad \dfrac{1}{4}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so in that case it would go to infinity

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1That's a good strategy with limits. If you're ever confused, just plug numbers in to see what happens.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Noooo :D An exponential can't go in the negative! :O

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or wait it would go to 0 not infinity right bc yeah what you just said haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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 yvalue) 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 yintersect 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
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.1Ok simmer down :O Let's look at the first one a sec.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large f(x)=Ca^x\]\[\large f(1)=6 \qquad \rightarrow \qquad 6=Ca^1\]\[f(3)=24 \qquad \rightarrow \qquad 24=Ca^{3}\]So this is where you got stuck on the first one?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1We 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large 4=a^2 \qquad \rightarrow \qquad a=2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok that makes sense = ) is that a property or a rule?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1It's this really fancy thing called "division" :3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or what i mean to say is how did you know to divide the points' equations hahahaha

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Well 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
 3 years ago
Best ResponseYou've already chosen the best response.1We could have also done substition to get the same result though :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0like solve for C in one of the problems and then plug it into the other equation?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok good i think i got it = ) thank you so much for all your help! = D
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