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Module 4 · L10 of 15 ~35 min ⚡ +95 XP available

Differentiating $a^x$

For bases other than $e$, differentiation introduces a correction factor of $\ln a$. Like tuning an instrument to a different key, changing the base requires adjusting by the natural log of that base. Understanding why reveals the full power of the exponential family.

Today's hook — You know $\dfrac{d}{dx}(e^x) = e^x$. But what about $\dfrac{d}{dx}(2^x)$? Is it just $2^x$? Or something else entirely? If it isn't $2^x$, what correction does the base $2$ introduce — and why does the natural logarithm show up in a differentiation formula?

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Recall — your gut answer first

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You know $\dfrac{d}{dx}(e^x) = e^x$. So what is $\dfrac{d}{dx}(2^x)$? Without using a formula — is it just $2^x$, or something else? Make a guess and explain your reasoning.

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The two moves

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Differentiating $a^x$ comes down to one key idea: rewrite $a^x$ using base $e$, then the chain rule produces the correction factor automatically.

Since $a = e^{\ln a}$, we have $a^x = e^{x \ln a}$. The chain rule then gives a factor of $\ln a$. When $a = e$, $\ln e = 1$ and the formula reduces to the familiar $e^x$ case.

$$\frac{d}{dx}(a^x) = a^x \ln a \qquad \frac{d}{dx}(a^{kx}) = ka^{kx} \ln a$$

The $\ln a$ factor is essential

$\dfrac{d}{dx}(2^x) = 2^x \ln 2 \approx 0.693 \cdot 2^x$. Forgetting $\ln a$ is the single most common error for this topic.

Why $\ln e = 1$ is the special case

$e^x$ gives $e^x \ln e = e^x \cdot 1 = e^x$. So the $e^x$ rule is just $a^x$ with $a = e$. One formula unifies both.

Fallback: rewrite as $e^{x \ln a}$

If you blank on the formula, write $a^x = e^{x \ln a}$ and apply chain rule: derivative $= e^{x \ln a} \cdot \ln a = a^x \ln a$.

What you'll master

Know

Key facts

$\dfrac{d}{dx}(a^x) = a^x \ln a$
$\dfrac{d}{dx}(a^{kx}) = ka^{kx} \ln a$
$a^x = e^{x \ln a}$ (rewriting technique)

Understand

Concepts

Where the $\ln a$ correction factor comes from
Why $e^x$ is the special case ($\ln e = 1$)
The link between $a^x$ and $e^{x \ln a}$

Can do

Skills

Differentiate $a^{kx}$ for any base $a$ and constant $k$
Apply product and quotient rules with $a^x$
Find stationary points and gradients for general exponentials

Key terms

General exponential derivative$\dfrac{d}{dx}(a^x) = a^x \ln a$ for any base $a > 0$, $a \neq 1$.

Logarithmic correction factorThe factor $\ln a$ that appears when differentiating $a^x$, converting the general base into the natural base $e$.

Rewriting technique$a^x = e^{x \ln a}$. Using this identity, the chain rule automatically produces the $\ln a$ factor.

Power rule vs exponential rule$\dfrac{d}{dx}(x^n) = nx^{n-1}$ (variable base) vs $\dfrac{d}{dx}(a^x) = a^x \ln a$ (variable exponent). Do not confuse these.

Where does $\ln a$ come from?

core concept

The derivation is elegant. Since $a = e^{\ln a}$, we can rewrite any exponential in base $e$:

$$a^x = e^{x \ln a} \quad \Rightarrow \quad \frac{d}{dx}(a^x) = \frac{d}{dx}(e^{x \ln a}) = \ln a \cdot e^{x \ln a} = a^x \ln a$$

The chain rule drops $\ln a$ out front (derivative of exponent $x \ln a$ with respect to $x$ is $\ln a$). When $a = e$, $\ln e = 1$ so the correction disappears entirely — confirming that $e^x$ is the natural special case.

Numeric check. $\dfrac{d}{dx}(2^x)\big|_{x=0} = 2^0 \ln 2 = \ln 2 \approx 0.693$. You can verify this with the limit definition: $\lim_{h \to 0} \dfrac{2^h - 1}{h} = \ln 2$. The $\ln 2$ emerges naturally from the limit — it is not an arbitrary constant.

Core formula: $\dfrac{d}{dx}(a^x) = a^x \ln a$ — the $\ln a$ factor is mandatory for $a \neq e$; Chain rule extension: $\dfrac{d}{dx}(a^{kx}) = k \ln a \cdot a^{kx}$ — both $k$ and $\ln a$ multiply

Pause — copy the rule $\dfrac{d}{dx}(a^x) = a^x \ln a$ and its chain rule extension $\dfrac{d}{dx}(a^{kx}) = k \ln a \cdot a^{kx}$ — both $k$ and $\ln a$ multiply — into your book.

Did you get this? True or false: $\dfrac{d}{dx}(3^x) = 3^x$ (without any $\ln 3$ factor).

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC $a^x$

Differentiate $y = 2^x$.

Band 3 Apply

$\dfrac{dy}{dx} = 2^x \ln 2$

Apply $\dfrac{d}{dx}(a^x) = a^x \ln a$ with $a = 2$. No chain rule needed — exponent is simply $x$.

PROBLEM 2 · CHAIN RULE WITH $a^{kx}$

Differentiate $y = 3^{2x}$.

Band 4 Apply

Let $u = 2x \Rightarrow u' = 2$

The exponent is $2x$, not just $x$ — chain rule needed.

PROBLEM 3 · PRODUCT RULE WITH $a^x$

Differentiate $y = x \cdot 5^x$.

Band 4 Apply

$u = x,\; v = 5^x \Rightarrow u' = 1,\; v' = 5^x \ln 5$

Product of linear function and general exponential. Differentiate each factor.

Quick check: What is $\dfrac{d}{dx}(3^{2x})$?

Common errors · the 3 traps that cost marks

Trap 01

Forgetting the $\ln a$ factor entirely

$\dfrac{d}{dx}(2^x) = 2^x \ln 2$, not $2^x$. The $\ln a$ factor is essential for all bases $a \neq e$. Writing $2^x$ as the derivative is the most common error for this topic — it costs full marks.

Trap 02

Applying the power rule $\dfrac{d}{dx}(a^x) = xa^{x-1}$

The power rule applies when the variable is in the base ($x^n$). Here the variable is in the exponent ($a^x$). These are fundamentally different. $\dfrac{d}{dx}(x^3) = 3x^2$ but $\dfrac{d}{dx}(3^x) = 3^x \ln 3$.

Trap 03

Including only one factor for $a^{kx}$

$\dfrac{d}{dx}(2^{3x}) = 3 \cdot 2^{3x} \ln 2$. Both the chain-rule factor ($3$) and the $\ln a$ factor ($\ln 2$) are required. Omitting either one loses marks.

Quick match: Which rule applies to each function?

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Differentiate $y = 4^x$.

Differentiate $y = 2^{3x}$.

Differentiate $y = x^2 \cdot 3^x$.

Differentiate $y = \dfrac{5^x}{x}$.

Find the gradient of $y = 10^x$ at $x = 0$.

Think-then-look: What is the derivative of $y = 4^x$? Write your answer before revealing.

Reveal answer

$\dfrac{dy}{dx} = 4^x \ln 4$. Apply $\dfrac{d}{dx}(a^x) = a^x \ln a$ with $a = 4$. Do not write $4x^3$ — that is the power rule for $x^4$, not $4^x$.

Revisit your thinking

Earlier you guessed the derivative of $2^x$. The answer is $2^x \ln 2$. The $\ln 2$ factor is the ‘correction’ that converts a general base $a$ into the natural base $e$ — because $a^x = e^{x \ln a}$, the chain rule drops a $\ln a$ out front. When $a = e$, $\ln e = 1$ and the correction vanishes, giving back $e^x$. This is not a coincidence — $e$ was chosen precisely because $\ln e = 1$.

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Odd one out: Three of these are correct. Which one is wrong?

Multiple choice

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Short answer

ApplyBand 42 marks

Q1. Differentiate general exponentials. Find $\dfrac{dy}{dx}$ for $y = 7^x + 7^{-x}$. Show working. (2 marks)

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ApplyBand 43 marks

Q2. Chain rule with non-linear exponent. Differentiate $y = 2^{x^2}$. Show full working. (3 marks)

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AnalyseBand 54 marks

Q3. Product with general exponential. Differentiate $y = (2x + 1) \cdot 3^x$ and find the value of $x$ where the gradient is zero. (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $4^x \ln 4$ · 2: $3\ln 2 \cdot 2^{3x}$ · 3: $3^x(2x \ln 3 + 2x/x + \ln 3 \cdot x^2)$… simplified: $3^x(2x + x^2 \ln 3)$ · 4: $\dfrac{5^x(x \ln 5 - 1)}{x^2}$ · 5: $10^0 \ln 10 = \ln 10 \approx 2.303$

Q1 (2 marks): $\dfrac{dy}{dx} = 7^x \ln 7 + 7^{-x} \cdot (-1) \cdot \ln 7 = \ln 7(7^x - 7^{-x})$ [2].

Q2 (3 marks): Rewrite as $2^{x^2}$, exponent $u = x^2$, $u' = 2x$ [0.5]. $\dfrac{dy}{dx} = 2^{x^2} \cdot \ln 2 \cdot 2x$ [1.5]. $= 2x \ln 2 \cdot 2^{x^2}$ [1].

Q3 (4 marks): Product rule: $\dfrac{dy}{dx} = 2 \cdot 3^x + (2x+1) \cdot 3^x \ln 3$ [1.5]. $= 3^x(2 + (2x+1)\ln 3)$ [1]. Set $= 0$: $3^x \neq 0$, so $2 + (2x+1)\ln 3 = 0$ [0.5]. $2x + 1 = -\dfrac{2}{\ln 3}$, $x = \dfrac{-1 - \frac{2}{\ln 3}}{2} \approx -1.41$ [1].

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← Lesson 9 · Differentiating $e^x$ Lesson 11 →

Module overview · Maths Advanced