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Derivative of aˣ (for any positive base a)

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.C (LO)
,
FUN‑3.C.1 (EK)
Sal finds the derivative of aˣ (for any positive base a) using the derivative of eˣ and the chain rule. He then differentiates 8⋅3ˣ.

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  • duskpin ultimate style avatar for user Rutwik Pasani
    Wouldn't the derivative of a^x be just x(a^(x-1)) according to the product rule?
    Or if not then why doesn't the product rule apply here?
    (69 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      No! This rule (actually called the power rule, not the product rule) only applies when the base is variable and the exponent is constant. I will assume that a is constant and the derivative is taken with respect to the variable x. In the expression a^x, the base is constant and the exponent is variable (instead of the other way around), so the power rule does not apply. The derivative of a^x with respect to x, assuming a is constant, is actually a^x * ln a.
      (132 votes)
  • orange juice squid orange style avatar for user Natsuko Nakagawa
    I don't understand why a = e^(ln(a)).
    (45 votes)
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    • aqualine ultimate style avatar for user Stefan van der Waal
      Let's start by drawing a partial logarithmic number line using e.
      |---|---|---|
      1 e e^2 e^3

      With every jump to the right, we multiply by e.

      ln(a) tells us how many jumps we have to make on this number line to get to a.
      So if a = e^3 ≈ 20.855, ln(a) = 3.

      If we raise e to the power we just calculated, 3, we get e^3, which is the a we started with.

      e^(ln(a)) is basically saying: first figure out how many jumps we have to make to get from 1 to a on the number line I drew at the beginning. After that, make that many jumps.
      (58 votes)
  • aqualine sapling style avatar for user Batchimeg D
    at wouldn't derivative of 8 equal to 0 ?
    (25 votes)
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    • duskpin sapling style avatar for user Vu
      Using the constant rule d/dx af(x) = a[d/dx f(x)]
      d/dx [8*3^x] = 8 [d/dx 3^x]
      So you don't differentiate 8 in this case.

      Had it been d/dx 8+3^x then you would use the sum rule, d/dx f(x) + g(x) = d/dx f(x) + d/dx g(x).

      d/dx 8 + 3^x = d/dx 8 + d/dx 3^x = 0 + ln(3) *3^x
      (44 votes)
  • leafers tree style avatar for user Marc
    How come when using the chain rule and taking the derivative ((ln a) * x) = ln a? Where does the x go?
    (18 votes)
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  • starky sapling style avatar for user 20leunge
    I don't see e^ln(a)*x as a composite function. If it can be expressed as f(g(x)), then what is f(x) and what is g(x)? And why is f'(g(x)) still e^ln(a)*x ?
    (21 votes)
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    • duskpin ultimate style avatar for user Michael Hopkins
      Ok, after going over this again and again, I believe I have worked this out.

      Let:
      f(u(x)) = e^u(x)
      u(x) =ln(a)*x
      G(x) = e^(ln(a)*x) = f(u(x))

      f'(u) = e^u (using the derivative of e rule)
      u'(x) = ln(a) (using constant multiple rule since ln(a) is a constant)

      so G'(x) = f'(u(x))*u'(x) (using the chain rule)

      substitute f'(u) and u'(x) as worked out above
      G'(x) = (e^u(x))*ln(a)

      substitute back in u(x)
      G'(x) = (e^(ln(a)*x))*ln(a)

      towards the beginning of the video, Sal determined that a = e^ln(a), so this can be substituted into the above equation of the the final answer of:

      G'(x) = (a^x)*ln(a)

      Hopefully they way I've written in out helps and doesn't cause any confusion!
      (21 votes)
  • leaf green style avatar for user Moly
    What will the formula be if a is not positive ? say d/dx[(-5)^x]
    (6 votes)
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    • leaf grey style avatar for user Alex
      The problem with (-5)^x is that it's only defined at a few select points, because values like (-5)^(1/2) are complex or imaginary, and ln of negative numbers is a bit complex (pun unintended). Thus, (-5)^x is undifferentiable over the reals; however, its derivative can still be found over the complex numbers as (-5)^x * (ln(5) + iπ).
      (8 votes)
  • duskpin ultimate style avatar for user jasminepandit
    What exactly does "a" represent in this video? Is it a variable (like x) or a constant (like e)?
    (4 votes)
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  • blobby blue style avatar for user mehrin farjana
    I know it's dumb to ask here but I don't know where to search so I'm just asking,
    is [y lnx] and [(lnx)y] the same thing?
    (3 votes)
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  • blobby green style avatar for user Marco Ramires
    How can he just assume that d/d(e^ln a) of [e^(ln a)x] is just equal to e^(ln a)x?
    (3 votes)
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  • aqualine ultimate style avatar for user Asmit Bhattacharya
    Why is ln(a).x the inside function? Shouldn't the inside function be e^ln(a) , the outside function raise it to the power x??
    (4 votes)
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Video transcript

- [Voiceover] What I want to do in this video is explore taking the derivatives of exponential functions. So we've already seen that the derivative with respect to x of e to the x is equal to e to x, which is a pretty amazing thing. One of the many things that makes e somewhat special. Though when you have an exponential with your base right over here as e, the derivative of it, the slope at any point, is equal to the value of that actual function. But now let's start exploring when we have other bases. Can we somehow figure out what is the derivative, what is the derivative with respect to x when we have a to the x, where a could be any number? Is there some way to figure this out? And maybe using our knowledge that the derivative of e to the x, is e to the x. Well can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base? Well, you could view a, you could view a as being equal to e. Let me write it this way. Well all right, a as being equal to e to the natural log of a. Now if this isn't obvious to you, I really want you to think about it. What is the natural log of a? The natural log of a is the power you need to raise e to, to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e too to get to a. Well then you're just going to get to a. So really think about this. Don't just accept this as a leap of faith. It should make sense to you. And it just comes out of really what a logarithm is. And so we can replace a with this whole expression here. If a is the same thing as e to the natural log of a, well then this is going to be, then this is going to be equal to the derivative with respect to x of e to the natural log, I keep writing la (laughs), to the natural log of a and then we're going to raise that to the xth power. We're going to raise that to the x power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color-coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a times x power. Times x power. And now we can use the chain rule to evaluate this derivative. So what we will do is we will first take the derivative of the outside function. So e to the natural log of a times x with respect to the inside function, with respect to natural log of a times x. And so, this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well natural log of a, it might not immediately jump out to you, but that's just going to be a number. So that's just going to be, so times the derivative. If it was the derivative of three x, it would just be three. If it's the derivative of natural log a times x, it's just going to be natural log of a. And so this is going to give us the natural log of a times e to the natural log of a. And I'm going to write it like this. Natural log of a to the x power. Well we've already seen this. This right over here is just a. So it all simplifies. It all simplifies to the natural log of a times a to the x, which is a pretty neat result. So if you're taking the derivative of e to the x, it's just going to be e to the x. If you're taking the derivative of a to the x, it's just going to be the natural log of a times a to the x. And so we can now use this result to actually take the derivatives of these types of expressions with bases other than e. So if I want to find the derivative with respect to x of eight times three to the x power, well what's that going to be? Well that's just going to be eight times and then the derivative of this right over here is going to be, based on what we just saw, it's going to be the natural log of our base, natural log of three times three to the x. Times three to the x. So it's equal to eight natural log of three times three to the x. Times three to the x power.