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bettyboop8904

  • 2 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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  1. incomplte
    • 2 years ago
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    You already have all the tools you need for differentiation. Do you have a specific example?

  2. zepdrix
    • 2 years ago
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    \[\large y=2^x\]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.

  3. bettyboop8904
    • 2 years ago
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    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?

  4. zepdrix
    • 2 years ago
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    oh lol :)

  5. zepdrix
    • 2 years ago
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    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.

  6. bettyboop8904
    • 2 years ago
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    lol

  7. incomplte
    • 2 years ago
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    lol @zepdrix has explained it better than i would

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

  9. bettyboop8904
    • 2 years ago
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    \[e^{2x}\div e^{4x ^{2}} + \cos(e ^{7x})\]

  10. bettyboop8904
    • 2 years ago
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    it approaches infinity?

  11. bettyboop8904
    • 2 years ago
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    that was me answering the post before i posted that big equation @zepdrix

  12. bettyboop8904
    • 2 years ago
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    i think i answered wrong lol

  13. zepdrix
    • 2 years ago
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    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!

  14. zepdrix
    • 2 years ago
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    What do we need to do with the big messy problem? Take it's derivative?

  15. bettyboop8904
    • 2 years ago
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    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?

  16. bettyboop8904
    • 2 years ago
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    And differentiate it

  17. bettyboop8904
    • 2 years ago
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    taking the derivative and differentiating are the same thing right?

  18. zepdrix
    • 2 years ago
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    Yes, just fancy words ^^

  19. bettyboop8904
    • 2 years ago
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    ok that's what i thought lol

  20. bettyboop8904
    • 2 years ago
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    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 = )

  21. zepdrix
    • 2 years ago
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    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.

  22. zepdrix
    • 2 years ago
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    Ok ok big question time c:

  23. bettyboop8904
    • 2 years ago
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    ok that makes sense = ) ty! and yes now to the big one lol

  24. zepdrix
    • 2 years ago
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    handy*

  25. bettyboop8904
    • 2 years ago
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    hold on lol cos is in the denominator with e to the 4x^2

  26. zepdrix
    • 2 years ago
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    Oh i see ^^

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

  28. zepdrix
    • 2 years ago
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    \[\large \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\]

  29. bettyboop8904
    • 2 years ago
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    right that's what they did with it = ) that's where I got confused, not with the actual rule but u'

  30. bettyboop8904
    • 2 years ago
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    and v'

  31. zepdrix
    • 2 years ago
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    \[\huge \left(e^{2x}\right)'=e^{2x}(2x)'\]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.

  32. zepdrix
    • 2 years ago
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    Remember how to do the chain rule? c:

  33. bettyboop8904
    • 2 years ago
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    yes

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

  35. bettyboop8904
    • 2 years ago
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    oh wow i think i just had a mental click lmfao

  36. zepdrix
    • 2 years ago
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    lol XD those are always fun.

  37. bettyboop8904
    • 2 years ago
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    hold on let me look at the problem and answer one more time lol

  38. bettyboop8904
    • 2 years ago
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    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

  39. zepdrix
    • 2 years ago
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    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:

  40. bettyboop8904
    • 2 years ago
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    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?

  41. bettyboop8904
    • 2 years ago
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    oh and then one more after that lol but i think it may be pretty easy haha

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

  43. zepdrix
    • 2 years ago
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    Since the base is LARGER than 1, and our exponent is positive, the value will explode.

  44. zepdrix
    • 2 years ago
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    What if you had,\[\large \lim_{x \rightarrow \infty} (.999)^x\]

  45. bettyboop8904
    • 2 years ago
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    so really big x values = really big y values

  46. zepdrix
    • 2 years ago
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    yah c: As x approaches infinity, y is approaching infinity. In that first case.

  47. bettyboop8904
    • 2 years ago
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    I used a calculator but the bigger x gets the closer y goes to 0 for the second part

  48. zepdrix
    • 2 years ago
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    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: \[\large \dfrac{1}{2}\cdot \dfrac{1}{2} \quad = \quad \dfrac{1}{4}\]

  49. zepdrix
    • 2 years ago
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    Yesss good good c:

  50. bettyboop8904
    • 2 years ago
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    correct = )

  51. bettyboop8904
    • 2 years ago
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    so in that case it would go to -infinity

  52. zepdrix
    • 2 years ago
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    That's a good strategy with limits. If you're ever confused, just plug numbers in to see what happens.

  53. zepdrix
    • 2 years ago
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    Noooo :D An exponential can't go in the negative! :O

  54. bettyboop8904
    • 2 years ago
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    or wait it would go to 0 not -infinity right bc yeah what you just said haha

  55. bettyboop8904
    • 2 years ago
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    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

  56. bettyboop8904
    • 2 years ago
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    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

  57. zepdrix
    • 2 years ago
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    Ok simmer down :O Let's look at the first one a sec.

  58. bettyboop8904
    • 2 years ago
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    lol

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

  60. bettyboop8904
    • 2 years ago
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    yes

  61. zepdrix
    • 2 years ago
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    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.\[\large \frac{24}{6}=\frac{\cancel{C}a^3}{\cancel{C}a^1}\]

  62. zepdrix
    • 2 years ago
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    \[\large 4=a^2 \qquad \rightarrow \qquad a=2\]

  63. bettyboop8904
    • 2 years ago
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    ok that makes sense = ) is that a property or a rule?

  64. zepdrix
    • 2 years ago
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    It's this really fancy thing called "division" :3

  65. zepdrix
    • 2 years ago
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    lolol

  66. bettyboop8904
    • 2 years ago
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    or what i mean to say is how did you know to divide the points' equations hahahaha

  67. zepdrix
    • 2 years ago
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    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

  68. zepdrix
    • 2 years ago
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    We could have also done substition to get the same result though :)

  69. bettyboop8904
    • 2 years ago
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    like solve for C in one of the problems and then plug it into the other equation?

  70. zepdrix
    • 2 years ago
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    Yah :)

  71. bettyboop8904
    • 2 years ago
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    ok good i think i got it = ) thank you so much for all your help! = D

  72. zepdrix
    • 2 years ago
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    yay team \c:/

  73. bettyboop8904
    • 2 years ago
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    lol

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