The idea

Differentiating exponentials and logarithms

The standard results d/dx eˣ = eˣ, d/dx eᵏˣ = k eᵏˣ, d/dx aˣ = aˣ ln a and d/dx ln x = 1/x, why e is the special base, and combining them with the chain, product and quotient rules.

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Pure · Differentiation

Differentiating exponentials and logarithms

The standard results d/dx eˣ = eˣ, d/dx eᵏˣ = k eᵏˣ, d/dx aˣ = aˣ ln a and d/dx ln x = 1/x, why e is the special base, and combining them with the chain, product and quotient rules.

Why it works

Proportional to its own height

Differentiate any exponential

y = a^x

from first principles and something striking happens. The gradient at

x

works out to

a^x

multiplied by a constant — the value of the gradient at

x = 0

\frac{d}{dx}a^x = a^x \times \Big(\lim_{h\to 0}\frac{a^h - 1}{h}\Big).

Read that back: the gradient of an exponential is the curve's own height, times a fixed number that depends only on the base.

The base that makes it exactly 1

So the curve

y = a^x

is, at every point, proportional to its own height. The only question is what that constant of proportionality is. For

a = 2

it is about

0.69

; for

a = 3

it is about

1.10

. Somewhere between

2

and

3

there is a base for which the constant is exactly $1$ — and that base is the number

e \approx 2.718

. This is the whole reason

e

is special:

\boxed{\frac{d}{dx}e^x = e^x.}

The exponential function

e^x

is its own derivative — the function that grows at a rate equal to its current size.

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