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## Differential Calculus

### Course: Differential Calculus > Unit 3

Lesson 2: More chain rule practice- Derivative of aˣ (for any positive base a)
- Derivative of logₐx (for any positive base a≠1)
- Derivatives of aˣ and logₐx
- Worked example: Derivative of 7^(x²-x) using the chain rule
- Worked example: Derivative of log₄(x²+x) using the chain rule
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Worked example: Derivative of ∜(x³+4x²+7) using the chain rule
- Chain rule capstone
- Proving the chain rule
- Derivative rules review

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# Derivative of aˣ (for any positive base a)

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.

## Want to join the conversation?

- 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?(70 votes)- 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.(140 votes)

- I don't understand why a = e^(ln(a)).(52 votes)
- 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.(65 votes)

- at5:00wouldn't derivative of 8 equal to 0 ?(27 votes)
- 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(46 votes)

- How come when using the chain rule and taking the derivative ((ln a) * x) = ln a? Where does the x go?(19 votes)
- Keep in mind that ln(a) is a constant; therefore this is the same as taking the derivative of
`cx`

where c = ln(a). The derivative would just be c in that case. Hope that helps.(46 votes)

- 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 ?(22 votes)
- 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!(23 votes)

- What will the formula be if a is not positive ? say d/dx[(-5)^x](7 votes)
- 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).(9 votes)

- What exactly does "a" represent in this video? Is it a variable (like x) or a constant (like e)?(5 votes)
- It's a constant, just one that we haven't chosen yet. We're assuming that some a has been decided on before we go taking derivatives and analyzing the function.(9 votes)

- How can he just assume that d/d(e^ln a) of [e^(ln a)x] is just equal to e^(ln a)x?(4 votes)
- We know that the derivative of 𝑒^𝑥, with respect to 𝑥, is 𝑒^𝑥.

Thereby, the derivative of 𝑒^(𝑥 ln 𝑎), with respect to 𝑥 ln 𝑎, is 𝑒^(𝑥 ln 𝑎).(6 votes)

- 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?(4 votes)- Multiplication is commutative, so y * ln(x) = ln(x) * y.(6 votes)

- 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??(6 votes)

## 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.