Single-Variable-Calculus · Unit 6 · Video 1 · Interactive Practice

The Derivative of $a^x$ and the Mystery Number $M(a)$

IKey Formulas

Formula	Role	What it contributes
$a^{x_1 + x_2} = a^{x_1}\,a^{x_2}$	Rule of exponents	Rewrites $a^{x+\Delta x}$ as $a^x a^{\Delta x}$ inside the difference quotient
$\dfrac{d}{dx}a^x = a^x \displaystyle\lim_{\Delta x \to 0} \dfrac{a^{\Delta x} - 1}{\Delta x}$	Factored definition	$x$ is fixed inside the limit, so $a^x$ comes outside
$M(a) \equiv \displaystyle\lim_{\Delta x \to 0} \dfrac{a^{\Delta x} - 1}{\Delta x}$	The mystery number	Depends only on the base $a$ ; equals the slope of $y = a^x$ at $(0, 1)$
$\dfrac{d}{dx}a^x = M(a)\,a^x$	The result	Slope at $x$ = slope at $0$ × height $a^x$

Key Insight: The rule of exponents reduces the derivative at every $x$ to a single number, the slope at $x = 0$ : $\left.\frac{d}{dx}a^x\right|_{x=0} = M(a)\,a^0 = M(a)$ . What we do not yet have is a way to compute $M(a)$ exactly.

IIVisualization 1 — $M(a)$ Is the Slope at $(0, 1)$

As $h \to 0$ , the secant slope $\frac{a^h - 1}{h}$ from $(0, 1)$ settles on one number: $M(a)$ .

Base

a

Shrink

h

IIIVisualization 2 — Factoring Out $a^x$

Adding exponents makes the rise over $[x, x+\Delta x]$ exactly $a^x$ times the rise over $[0, \Delta x]$ .

Step 1

Step 1 — The definition of the derivative

\frac{d}{dx}a^x = \lim_{\Delta x \to 0}\frac{a^{x+\Delta x} - a^x}{\Delta x}

\text{here: } \frac{2^{1.5} - 2^{1}}{0.5}

Step 2 — Add exponents, multiply powers

a^{x+\Delta x} = a^x\,a^{\Delta x} \qquad 2^{1.5} = 2^{1}\cdot 2^{0.5}

= \lim_{\Delta x \to 0}\frac{a^x a^{\Delta x} - a^x}{\Delta x}

Step 3 — Factor out $a^x$

= \lim_{\Delta x \to 0} a^x\,\frac{a^{\Delta x} - 1}{\Delta x}

2^{1.5} - 2^{1} = 2^{1}\,\big(2^{0.5} - 1\big)

Step 4 — $x$ fixed, $\Delta x$ moves

\frac{d}{dx}a^x = a^x \lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x}

The remaining limit contains no

x

at all.

Step 5 — Name the limit

M(a) \equiv \lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x}, \qquad \frac{d}{dx}a^x = M(a)\,a^x

\left.\frac{d}{dx}2^x\right|_{x=1} = 2\,M(2)

IVVisualization 3 — Slope Equals $M(a)$ Times Height

At every point of $y = a^x$ , slope divided by height is the same number, $M(a)$ .

Base

a

Point

x

1.5

💡 For $\sin x$ the slope at zero, $\sin'(0) = 1$ , came from measuring angles in radians. For $a^x$ there is no such method yet; the next video chooses the one base whose slope at zero is exactly $1$ .

VQuiz Questions

Problem 1 · Slope Away From Zero

Given: $\dfrac{d}{dx}a^x = M(a)\,a^x$ . Find the slope of $y = 2^x$ at $x = 3$ , in terms of $M(2)$ .

✅ Correct! Slope at

x = 3

is the slope at

0

times the height:

M(2)\cdot 2^3 = 8\,M(2)

❌ That is the power rule.

x\cdot 2^{x-1}

treats

2^x

x^n

, but here the exponent is the variable and the base is fixed.

❌ That is the slope at $x = 0$ only. Away from zero it is multiplied by the height

2^x

❌ Multiply by the height, not by $x$ . The factor is

a^x = 2^3

, not

3

Show solution

\left.\frac{d}{dx}2^x\right|_{x=3} = M(2)\cdot 2^{3} = 8\,M(2)

The slope anywhere equals the slope at zero, $M(2)$ , multiplied by the height of the curve, $2^3 = 8$ .

Problem 2 · A Negative Exponent

Find the slope of $y = 2^x$ at $x = -1$ , in terms of $M(2)$ .

✅ Correct! The height at

x = -1

2^{-1} = \tfrac{1}{2}

, so the slope is

\tfrac{1}{2}M(2)

— still positive, since the curve rises everywhere.

❌ A negative exponent gives a reciprocal, not a negative.

2^{-1}\cdot 2^{1} = 2^0 = 1

forces

2^{-1} = \tfrac{1}{2}

❌ That applies the power rule to $2^x$ . The exponent is the variable, so

\frac{d}{dx}2^x = M(2)\,2^x

❌ $M(2)$ is the slope at $x = 0$ . At

x = -1

it is scaled by the height

2^{-1}

Show solution

From $2^{-1}\cdot 2^{1} = 2^{0} = 1$ we get $2^{-1} = \tfrac{1}{2}$ . Then

\left.\frac{d}{dx}2^x\right|_{x=-1} = M(2)\cdot 2^{-1} = \tfrac{1}{2}\,M(2).

The slope is positive (the height $a^x$ is always positive) and smaller than at $x = 0$ , because the curve is lower there.

Problem 3 · Find the Base From a Slope Ratio

Given: for some base $a > 0$ , the slope of $y = a^x$ at $x = 2$ is $9$ times its slope at $x = 0$ . Find $a$ , then compare the slope at $x = -2$ with the slope at $x = 0$ .

What is the base $a$ ?

The slope at $x = -2$ equals the slope at $x = 0$ times…

✅ Correct! The ratio of slopes is the ratio of heights:

a^2 = 9

gives

a = 3

, and at

x = -2

the factor is

3^{-2} = \tfrac{1}{9}

❌ Check the base. Slope at

x

÷ slope at

0

= \dfrac{M(a)\,a^x}{M(a)} = a^x

, so here

a^2 = 9

❌ The ratio $9$ is $a^2$ , not $a$ . The factor at

x = 2

is the height

a^2

❌ The factor is $a^x$ , not $a\cdot x$ . Solve

a^2 = 9

instead of

2a = 9

❌ Not quite. The factor at

x = -2

is the height

3^{-2}

❌ A negative exponent gives a reciprocal, not a negative. The height

3^{-2}

is positive, and so is every slope of

y = 3^x

Show solution

Since $\frac{d}{dx}a^x = M(a)\,a^x$ , the slope at $x$ divided by the slope at $0$ is

\frac{M(a)\,a^{x}}{M(a)\,a^{0}} = a^{x}.

At $x = 2$ : $a^2 = 9$ , and with $a > 0$ this gives $a = 3$ .

At $x = -2$ : the factor is $3^{-2} = \dfrac{1}{9}$ , so the slope there is $\tfrac{1}{9}M(3)$ .

Problem 4 · How $M(4)$ Relates to $M(2)$

Given: $4^x = (2^x)^2$ (a power of a power). Differentiate both sides — the right side with the chain rule — and find the relation between $M(4)$ and $M(2)$ .

✅ Correct! Doubling the exponent doubles the slope at zero:

M(4) = 2\,M(2)

❌ Squaring the function does not square its slope. The chain rule gives

2(2^x)\cdot M(2)\,2^x

, which is linear in

M(2)

❌ $M(a)$ depends on the base.

y = 4^x

is steeper than

y = 2^x

through the same point

(0, 1)

, so its slope there is larger.

❌ Not quite. Compare the two derivatives of

4^x

M(4)\,4^x

and

2(2^x)\cdot M(2)\,2^x = 2M(2)\,4^x

Show solution

Left side: $\dfrac{d}{dx}4^x = M(4)\,4^x$ .

Right side (chain rule, outer function $u^2$ , inner $u = 2^x$ ):

\frac{d}{dx}(2^x)^2 = 2\,(2^x)\cdot M(2)\,2^x = 2\,M(2)\,(2^x)^2 = 2\,M(2)\,4^x.

The two sides are equal for every $x$ , so $M(4)\,4^x = 2\,M(2)\,4^x$ and $M(4) = 2\,M(2)$ .

Check from the definition: $\dfrac{4^h - 1}{h} = \dfrac{(2^h - 1)(2^h + 1)}{h}$ , and as $h \to 0$ the factor $2^h + 1 \to 2$ , giving $M(4) = 2\,M(2)$ again.

Solved: 0 / 4

The Derivative of axa^xax and the Mystery Number M(a)M(a)M(a)

IKey Formulas

IIVisualization 1 — M(a)M(a)M(a) Is the Slope at (0,1)(0, 1)(0,1)

IIIVisualization 2 — Factoring Out axa^xax

IVVisualization 3 — Slope Equals M(a)M(a)M(a) Times Height

VQuiz Questions

The Derivative of $a^x$ and the Mystery Number $M(a)$

IIVisualization 1 — $M(a)$ Is the Slope at $(0, 1)$

IIIVisualization 2 — Factoring Out $a^x$

IVVisualization 3 — Slope Equals $M(a)$ Times Height