Single-Variable-Calculus · Unit 6 · Video 3 · Interactive Practice

$M(a) = \ln a$ : Differentiating $a^x$ and $x^x$

IKey Formulas

Formula	Name	Where it comes from
$a^x = e^{x\ln a}$	Convert to base $e$	$a = e^{\ln a}$ , then a power of a power multiplies exponents
$\dfrac{d}{dx}\,a^x = (\ln a)\,a^x$	So $M(a) = \ln a$	Chain rule on $e^{(\ln a)x}$ , exactly like $e^{3x}$
$(\ln u)' = \dfrac{u'}{u}$	Logarithmic differentiation	Chain rule with $\frac{d}{du}\ln u = \frac{1}{u}$
$\dfrac{d}{dx}\,x^x = x^x(1 + \ln x)$	Variable base and exponent	$\ln v = x\ln x$ , then the product rule

Key Insight: In $e^{(\ln a)x}$ the letters $\ln a$ name a fixed number, so they sit where the $3$ sits in $e^{3x}$ and the chain rule brings them down: $\frac{d}{dx}e^{3x} = 3e^{3x}$ , $\ \frac{d}{dx}e^{(\ln a)x} = (\ln a)\,e^{(\ln a)x}$ .

IIVisualization 1 — The Mystery Number on the Graph

The slope of $a^x$ is its height times one fixed number, $M(a) = \ln a$ — the slope at $x = 0$ .

Base a 2.00 Point x 1.00

💡 Only $a = e$ gives slope $1$ at $x = 0$ , since $\ln e = 1$ ; bases below $1$ have $\ln a < 0$ and a falling graph.

IIIVisualization 2 — Logarithmic Differentiation of $x^x$

Taking logs turns $x^x$ into the product $x\ln x$ , which the product rule can handle.

Step 1

Step 1 — Base and exponent both vary

v = x^x,\qquad x > 0

Step 2 — Take logs

\ln v = x\ln x

Step 3 — Product rule; left side is $v'/v$

\frac{v'}{v} = 1\cdot\ln x + x\cdot\frac{1}{x} = 1 + \ln x

Step 4 — Multiply through by $v$

v' = v\,(1 + \ln x) = x^x(1 + \ln x)

💡 Write $1 + \ln x$ with the $1$ first: it is not $\ln(x + 1)$ .

IVVisualization 3 — Two Pieces of One Slope

The slope of $x^x$ is the exponent piece $x^x\ln x$ plus the base piece $x^x$ — neither rule alone.

Point x 1.30

Product rule, piece by piece

(x\ln x)' = \underbrace{1\cdot\ln x}_{\text{from the exponent}} + \underbrace{x\cdot\tfrac{1}{x}}_{\text{from the base}}

Multiply by $x^x$

\frac{d}{dx}\,x^x = \underbrace{x^x\ln x}_{\text{exponential rule alone}} + \underbrace{x^x}_{\text{power rule alone}}

Check by base $e$

\frac{d}{dx}\,e^{x\ln x} = e^{x\ln x}\,(x\ln x)' = x^x(1 + \ln x)

VQuiz Questions

Problem 1 · Differentiate an Exponential

Find: $\dfrac{d}{dx}\,3^x$ .

✅ Correct!

M(3) = \ln 3 \approx 1.099

, so

3^x

grows at about

1.1

times its own height.

❌ That is the power rule. The power rule needs a fixed exponent; in

3^x

the exponent varies and the base is fixed.

❌ Only base $e$ reproduces itself.

M(3) = \ln 3

, and

\ln 3 \neq 1

❌ Not quite. Write

3^x = e^{x\ln 3}

; the chain rule multiplies by the constant

\ln 3

, it does not divide.

Show solution

Convert to base $e$ : $3 = e^{\ln 3}$ , so $3^x = e^{(\ln 3)x}$ . Since $\ln 3$ is a constant, the chain rule gives

\frac{d}{dx}\,e^{(\ln 3)x} = (\ln 3)\,e^{(\ln 3)x} = (\ln 3)\,3^x.

So $\frac{d}{dx}3^x = (\ln 3)\,3^x$ , and $M(3) = \ln 3 \approx 1.099$ .

Problem 2 · Slope of x^x at x = 2

Find: the slope of $y = x^x$ at $x = 2$ .

✅ Correct!

4(1 + \ln 2) \approx 6.77

: the base piece

4

plus the exponent piece

4\ln 2 \approx 2.77

❌ Power rule alone.

x\cdot x^{x-1} = x^x = 4

treats the exponent as fixed, so it keeps only the base piece.

❌ Exponential rule alone.

x^x\ln x = 4\ln 2

treats the base as fixed, so it keeps only the exponent piece.

❌ That reads $1 + \ln x$ as $\ln(x + 1)$ . At

x = 2

the factor is

1 + \ln 2

, not

\ln 3

❌ Not quite. Use

\frac{d}{dx}x^x = x^x(1 + \ln x)

x = 2

Show solution

Let $v = x^x$ . Then $\ln v = x\ln x$ , and the product rule gives

\frac{v'}{v} = 1\cdot\ln x + x\cdot\frac{1}{x} = 1 + \ln x \quad\Longrightarrow\quad v' = x^x(1 + \ln x).

At $x = 2$ : $v' = 2^2(1 + \ln 2) = 4(1 + \ln 2) \approx 4(1.693) \approx 6.77$ .

Problem 3 · Where x^(1/x) Levels Off

Given: $y = x^{1/x}$ for $x > 0$ . Use logarithmic differentiation to find $y'/y$ and the $x$ where the tangent line is horizontal.

What is y′/y?

Where is the tangent line horizontal?

✅ Correct!

y' = 0

exactly when

1 - \ln x = 0

, at

x = e

, where the height is

e^{1/e} \approx 1.445

❌ Check the order in the quotient rule.

\left(\frac{\ln x}{x}\right)' = \frac{\frac{1}{x}\cdot x - \ln x\cdot 1}{x^2}

: the

1

comes first.

❌ Take logs first.

\ln y = \frac{\ln x}{x}

; differentiate that with the quotient rule, keeping both terms.

❌ That solves $1 + \ln x = 0$ . That was the

x^x

factor; here the factor is

1 - \ln x

❌ Not quite. Since

y > 0

and

x^2 > 0

y' = 0

exactly when

1 - \ln x = 0

Show solution

Take logs: $\ln y = \frac{1}{x}\ln x = \frac{\ln x}{x}$ . The quotient rule gives

\frac{y'}{y} = \frac{\frac{1}{x}\cdot x - \ln x\cdot 1}{x^2} = \frac{1 - \ln x}{x^2}.

Multiply through by $y$ : $y' = x^{1/x}\,\dfrac{1 - \ln x}{x^2}$ .

Both $x^{1/x}$ and $x^2$ are positive, so $y' = 0$ only when $\ln x = 1$ , that is $x = e$ . The height there is $e^{1/e} \approx 1.445$ .

Problem 4 · A Variable Exponent Inside

Find: $\dfrac{d}{dx}\,2^{x^2}$ .

✅ Correct! Base

e

turns it into

e^{(\ln 2)x^2}

, and the chain rule brings down

(\ln 2)\cdot 2x

❌ Power rule on a fixed base. The base

2

is fixed and the exponent

x^2

varies, so the power rule does not apply.

❌ Missing the inner derivative. The exponent is

x^2

, not

x

, so the chain rule also multiplies by

2x

❌ Missing $\ln 2$ .

2^u

has derivative

(\ln 2)\,2^u

with respect to

u

, not

2^u

❌ Not quite. Write

2^{x^2} = e^{(\ln 2)x^2}

and differentiate the exponent.

Show solution

Convert to base $e$ : $2^{x^2} = e^{(\ln 2)\,x^2}$ . The chain rule multiplies by the derivative of the exponent, $(\ln 2)\cdot 2x$ :

\frac{d}{dx}\,e^{(\ln 2)x^2} = (\ln 2)(2x)\,e^{(\ln 2)x^2} = (2\ln 2)\,x\,2^{x^2}.

Check by logs: $\ln y = x^2\ln 2$ , so $\frac{y'}{y} = 2x\ln 2$ and $y' = (2\ln 2)\,x\,2^{x^2}$ .

Solved: 0 / 4

M(a)=ln⁡aM(a) = \ln aM(a)=lna: Differentiating axa^xax and xxx^xxx

IKey Formulas

IIVisualization 1 — The Mystery Number on the Graph

IIIVisualization 2 — Logarithmic Differentiation of xxx^xxx

IVVisualization 3 — Two Pieces of One Slope

VQuiz Questions

$M(a) = \ln a$ : Differentiating $a^x$ and $x^x$

IIIVisualization 2 — Logarithmic Differentiation of $x^x$