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bettyboop8904

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

  • one year ago
  • one year ago

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

    • one year ago
  2. zepdrix
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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.

    • one year ago
  3. bettyboop8904
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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?

    • one year ago
  4. zepdrix
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    oh lol :)

    • one year ago
  5. zepdrix
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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.

    • one year ago
  6. bettyboop8904
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    lol

    • one year ago
  7. incomplte
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    lol @zepdrix has explained it better than i would

    • one year ago
  8. zepdrix
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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?

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

    • one year ago
  10. bettyboop8904
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    it approaches infinity?

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

    • one year ago
  12. bettyboop8904
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    i think i answered wrong lol

    • one year ago
  13. zepdrix
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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!

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

    • one year ago
  15. bettyboop8904
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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?

    • one year ago
  16. bettyboop8904
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    And differentiate it

    • one year ago
  17. bettyboop8904
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    taking the derivative and differentiating are the same thing right?

    • one year ago
  18. zepdrix
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    Yes, just fancy words ^^

    • one year ago
  19. bettyboop8904
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    ok that's what i thought lol

    • one year ago
  20. bettyboop8904
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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 = )

    • one year ago
  21. zepdrix
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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.

    • one year ago
  22. zepdrix
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    Ok ok big question time c:

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

    • one year ago
  24. zepdrix
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    handy*

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

    • one year ago
  26. zepdrix
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    Oh i see ^^

    • one year ago
  27. zepdrix
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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.

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

    • one year ago
  29. bettyboop8904
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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'

    • one year ago
  30. bettyboop8904
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    and v'

    • one year ago
  31. zepdrix
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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.

    • one year ago
  32. zepdrix
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    Remember how to do the chain rule? c:

    • one year ago
  33. bettyboop8904
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    yes

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

    • one year ago
  35. bettyboop8904
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    oh wow i think i just had a mental click lmfao

    • one year ago
  36. zepdrix
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    lol XD those are always fun.

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

    • one year ago
  38. bettyboop8904
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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

    • one year ago
  39. zepdrix
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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:

    • one year ago
  40. bettyboop8904
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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?

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

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

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

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

    • one year ago
  45. bettyboop8904
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    so really big x values = really big y values

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

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

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

    • one year ago
  49. zepdrix
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    Yesss good good c:

    • one year ago
  50. bettyboop8904
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    correct = )

    • one year ago
  51. bettyboop8904
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    so in that case it would go to -infinity

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

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

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

    • one year ago
  55. bettyboop8904
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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

    • one year ago
  56. bettyboop8904
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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

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

    • one year ago
  58. bettyboop8904
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    lol

    • one year ago
  59. zepdrix
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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?

    • one year ago
  60. bettyboop8904
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    yes

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

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

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

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

    • one year ago
  65. zepdrix
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    lolol

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

    • one year ago
  67. zepdrix
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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

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

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

    • one year ago
  70. zepdrix
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    Yah :)

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

    • one year ago
  72. zepdrix
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    yay team \c:/

    • one year ago
  73. bettyboop8904
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    lol

    • one year ago
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