Exponential and logarithmic functions

Among every function you have differentiated so far — polynomials, trig ratios, roots — one function stands apart: $e^x$ is its own derivative. Differentiate it and nothing changes. That single fact makes $e^x$ and its inverse partner $\ln x$ the backbone of the rest of calculus, from logarithmic differentiation to almost every integration trick you will meet next. This topic sits inside Continuity and Differentiability, right after implicit differentiation, and leans hard on the chain, product and quotient rules you already own. The goal here is narrow and concrete: learn the two base derivatives, learn how the chain rule dresses them up as $e^{g(x)}$ and $\ln[g(x)]$ , and learn the general-base cousins $a^x$ and $\log_a x$ that bring in $\ln a$ . Master this and logarithmic differentiation — next up — becomes almost mechanical.

The core insight

The exponential function $e^x$ is the unique function equal to its own derivative, and this one fact — together with the chain rule — generates every exponential and logarithmic derivative rule you need.

The core idea

The two foundational results for this topic are the base derivatives of $e^x$ and $\ln x$ :

\frac{d}{dx}(e^x) = e^x, \qquad \frac{d}{dx}(\ln x) = \frac{1}{x} \quad (x>0)

where: $e \approx 2.71828$ is Euler's number, the unique base for which the exponential's growth rate at $x=0$ equals $1$ ; $\ln x$ (also written $\log_e x$ ) is the natural logarithm, defined only where $x>0$ , so its derivative formula inherits that same restricted domain. No other elementary function reproduces itself under differentiation — this self-replicating growth is exactly what makes $e^x$ special, and it is why $e$ is chosen as the "natural" base in the first place. Real problems rarely hand you a bare $x$ inside the exponential or the log — they hand you a composite function. Bolting on the chain rule turns the two base results into the forms you will actually use:

\frac{d}{dx}\left(e^{g(x)}\right) = e^{g(x)}\cdot g'(x), \qquad \frac{d}{dx}\left(\ln[g(x)]\right) = \frac{g'(x)}{g(x)} \quad (g(x)>0)

where: $g(x)$ is any differentiable inner function and $g'(x)$ its derivative; the outer derivative ( $e^u$ or $1/u$ ) is multiplied by the inner derivative, exactly as the chain rule demands. Two auxiliary results extend the idea to any positive base $a \neq 1$ , obtained by rewriting $a^x = e^{x\ln a}$ and applying the chain rule above:

\frac{d}{dx}(a^x) = a^x \ln a, \qquad \frac{d}{dx}(\log_a x) = \frac{1}{x\ln a} \quad (x>0,\ a>0,\ a\neq 1)

where: $\ln a$ is a fixed numerical constant for a given base $a$ (not generally $1$ — it equals $1$ only when $a=e$ ). This is exactly why $e^x$ sits above every other exponential: for any other base the derivative carries an extra factor of $\ln a$ , and only $a=e$ makes that factor vanish to $1$ . It helps to notice that $e^x$ and $\ln x$ are inverse functions of each other, so their graphs are mirror images across the line $y=x$ — one climbs steeply, the other climbs gently, and reflecting one across that diagonal produces the other exactly. This inverse relationship is also why differentiating $y=\ln x$ can be checked by writing $x=e^y$ and implicitly differentiating: $1 = e^y\dfrac{dy}{dx}$ , so $\dfrac{dy}{dx}=\dfrac{1}{e^y}=\dfrac{1}{x}$ , matching the stated formula and confirming the two base rules are consistent with each other rather than two unrelated facts to memorise separately.

Figure

A curve for e to the x rising steeply and its mirror image the natural log curve rising slowly, reflected across the line y equals x, with a tangent line on the exponential curve matching the height of the curve at that point

Derivation of the exponential rule

Derivation AWhy e to the x is its own derivative

The goal is to see, at the level expected in this syllabus, why differentiating $e^x$ returns $e^x$ exactly — not approximately, not up to a constant — using only the definition of $e$ as the base whose exponential grows at unit rate at the origin, and the exponent law $e^{h}-1 \approx h$ for small $h$ .

Step 1. Start from the definition of the derivative as a limit: $\dfrac{d}{dx}(e^x) = \displaystyle\lim_{h\to 0} \frac{e^{x+h}-e^x}{h}$ .

Step 2. Split the exponent using the law $e^{x+h}=e^x\cdot e^h$ , then factor out the common term $e^x$ , which does not depend on $h$ : $\dfrac{e^{x+h}-e^x}{h} = \dfrac{e^x e^h - e^x}{h} = e^x\cdot\dfrac{e^h-1}{h}$ .

Step 3. Evaluate the remaining limit $\displaystyle\lim_{h\to 0}\dfrac{e^h-1}{h}$ . This is precisely the defining property of the number $e$ : it is chosen as the unique base for which this limit equals $1$ (equivalently, the exponential's instantaneous growth rate at $x=0$ is exactly $1$ ). Any other base $a$ gives $\lim_{h\to 0}\frac{a^h-1}{h}=\ln a \neq 1$ in general.

Step 4. Substitute this limit value back in: $\dfrac{d}{dx}(e^x) = e^x\cdot\displaystyle\lim_{h\to 0}\frac{e^h-1}{h} = e^x \cdot 1 = e^x$ . The factoring in Step 2 is what makes the result so clean — the constant multiple $e^x$ simply rides along untouched, and the entire content of the derivative sits in that one special limit equal to $1$ .

Worked example

Differentiate $y = e^{3x^2+1}$ with respect to $x$ , and also differentiate $y = \ln(x^2+3x)$ , evaluating its derivative at $x=1$ .

Step 1. For $y=e^{3x^2+1}$ , recognise the composite structure: outer function $e^u$ , inner function $u=3x^2+1$ .

Step 2. Differentiate the outer function with respect to $u$ : $\dfrac{dy}{du}=e^u=e^{3x^2+1}$ , using $\frac{d}{du}e^u=e^u$ from the box above.

Step 3. Differentiate the inner function: $\dfrac{du}{dx}=6x$ . Apply the chain rule: $\dfrac{dy}{dx}=e^{3x^2+1}\cdot 6x = 6x\,e^{3x^2+1}$ .

Step 4. Now for $y=\ln(x^2+3x)$ , the outer function is $\ln u$ with inner $u=x^2+3x$ , so $\dfrac{dy}{du}=\dfrac{1}{x^2+3x}$ and $\dfrac{du}{dx}=2x+3$ , giving $\dfrac{dy}{dx}=\dfrac{2x+3}{x^2+3x}$ .

Step 5. Evaluate at $x=1$ , but first check the domain: $x^2+3x=1+3=4>0$ , so $\ln(x^2+3x)$ and its derivative are defined there. Then $\left.\dfrac{dy}{dx}\right|_{x=1}=\dfrac{2(1)+3}{1^2+3(1)}=\dfrac{5}{4}$ .

Final answers: $\dfrac{dy}{dx}=6x\,e^{3x^2+1}$ for the first function, and $\dfrac{dy}{dx}=\dfrac{2x+3}{x^2+3x}$ , giving $\dfrac{5}{4}$ at $x=1$ , for the second.

Competitive stretch

If instead you needed the derivative at $x=-1$ , checking the domain first would stop you cold: $x^2+3x = 1-3=-2<0$ there, so $\ln(x^2+3x)$ is not even defined at $x=-1$ , and no derivative question could sensibly be asked at that point — always test the sign of $g(x)$ before differentiating $\ln[g(x)]$ at a specific value.

Common traps

Confusing $\frac{d}{dx}(e^x)=e^x$ with the derivative of $x^e$ . These are different families: $x^e$ uses the power rule, giving $ex^{e-1}$ , while $e^x$ reproduces itself.
Dropping the chain-rule factor on $\ln[g(x)]$ . Writing $\frac{1}{g(x)}$ instead of the correct $\frac{g'(x)}{g(x)}$ forgets to multiply by the inner derivative.
Ignoring the domain restriction $x>0$ . Differentiating $\ln[g(x)]$ silently assumes $g(x)>0$ on the interval in question; skipping this check can lead to nonsense at specific points.
Mixing up $a^x$ and $x^a$ . $\frac{d}{dx}(a^x)=a^x\ln a$ is the exponential rule with constant base $a$ ; $\frac{d}{dx}(x^a)=ax^{a-1}$ is the power rule with constant exponent $a$ — superficially similar, structurally unrelated.
Assuming $\ln a$ always equals $1$ . That holds only for $a=e$ ; for $a=2$ or $a=10$ , $\ln a$ is a genuine numerical constant that must be carried through the calculation.

Ask the tutor: if the difference between differentiating $e^x$ , $x^e$ , and $a^x$ still feels blurry, ask for a quick side-by-side worked comparison with one domain-check example thrown in.

Why is e chosen as the base for the natural logarithm and not, say, 10?

Because $e$ is the unique base for which $\frac{d}{dx}(e^x)=e^x$ exactly, with no extra multiplying constant. Every other base $a$ gives $\frac{d}{dx}(a^x)=a^x\ln a$ , carrying an awkward factor of $\ln a$ through every differentiation and integration. Choosing $a=e$ makes that factor equal to $1$ , which is precisely why calculus is built around $e$ rather than $10$ — it is the base that makes the rules cleanest, not an arbitrary convention.

Do I always need the chain rule when differentiating

e^x

\ln x

Only when the exponent or the argument is itself a function of $x$ other than $x$ alone. Plain $e^x$ or $\ln x$ needs no chain rule — the base formulas apply directly. But the moment you see $e^{3x^2+1}$ or $\ln(x^2+3x)$ , the "inside" is a function of $x$ , and you must multiply by its derivative, exactly as shown in the worked example above. Spotting the composite structure first is the whole trick.

How is the derivative of

\log_a x

different from the derivative of

\ln x

They differ by a constant factor of $\ln a$ in the denominator: $\frac{d}{dx}(\log_a x) = \frac{1}{x\ln a}$ , compared with $\frac{d}{dx}(\ln x) = \frac{1}{x}$ . Since $\ln e = 1$ , setting $a=e$ in the general formula collapses it exactly to the natural-log case — so $\ln x$ is not a separate rule to memorise, it is simply the $a=e$ special case of $\log_a x$ where the extra factor disappears.