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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 27: Exponential functions differentiation

# Derivative of aˣ (for any positive base a)

Let's dive deeper into the fascinating world of derivatives, specifically focusing on the derivative of aˣ for any positive base a. Using the derivative of eˣ and the chain rule, we unravel the mystery behind differentiating exponential functions. We then apply our newfound knowledge to differentiate the expression 8⋅3ˣ.

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## 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? •   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.
• I don't understand why a = e^(ln(a)). •   Let's start by drawing a partial logarithmic number line using e.
|---|---|---|1   e  e^2 e^3

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.
• at wouldn't derivative of 8 equal to 0 ? •  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
• How come when using the chain rule and taking the derivative ((ln a) * x) = ln a? Where does the x go? •  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.
• 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 ? • 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!
• What will the formula be if a is not positive ? say d/dx[(-5)^x] • 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).
• What exactly does "a" represent in this video? Is it a variable (like x) or a constant (like e)? • How can he just assume that d/d(e^ln a) of [e^(ln a)x] is just equal to e^(ln a)x?   