Derivatives of Basic Functions

You already know how to compute a derivative from the definition: write down the difference quotient \frac{f(x+h) - f(x)}{h}, simplify, take h \to 0. You used this to show that the derivative of x^2 is 2x, and the derivative of 1/x is -1/x^2.

But nobody wants to expand (x+h)^{17} by hand just to differentiate x^{17}. There must be a pattern.

There is. This article derives the four fundamental derivative formulas — the ones that every other derivative in calculus ultimately rests on. Each proof goes from the limit definition to the final formula, with every step shown. Once you have these, the rules of differentiation will let you combine them to differentiate anything.

The constant rule

Start with the simplest possible function: f(x) = c, where c is a constant. A horizontal line.

This says exactly what you would expect: a flat line has zero slope everywhere. The function does not change, so its rate of change is zero. The graph of f(x) = c is a horizontal line, and horizontal lines have slope zero. The derivative simply encodes this geometrically obvious fact.

Worth noting: the constant can be anything — f(x) = 5, f(x) = -\pi, f(x) = \sqrt{2}. The derivative is always zero because the function never moves. This is the base case that every other derivative formula builds on.

The power rule

This is the single most-used derivative formula. It handles x^2, x^3, x^{100}, x^{1/2}, x^{-1}, and x^n for any real number n.

The case of positive integers

Start with f(x) = x^n where n is a positive integer. You need to expand (x+h)^n. The binomial theorem gives

Every term after the first two contains h^2 or higher powers of h. So

Divide by h:

Every term except the first contains at least one factor of h. As h \to 0, all those terms vanish, leaving

That is the power rule for positive integer exponents, proved in four lines from the binomial theorem.

Check it against the cases you already know. For n = 2: f'(x) = 2x^{2-1} = 2x, which matches the derivative you computed from the definition in the derivative article. For n = 3: f'(x) = 3x^2. The derivative of x^3 is 3x^2 — you can verify by expanding (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3, subtracting x^3, dividing by h, and taking the limit, which gives 3x^2 + 3xh + h^2 \to 3x^2. The pattern holds.

An alternative proof using factoring

There is a second proof of the power rule for positive integers that does not use the binomial theorem at all. It uses the algebraic identity

Set a = x + h and b = x:

The factor (x + h - x) simplifies to h, which cancels with the h in the denominator:

This is a sum of n terms. As h \to 0, every (x+h) becomes x, so each term becomes x^{n-1}. The limit is n \cdot x^{n-1}.

This proof is elegant because it avoids the binomial expansion entirely. It works by recognising that a^n - b^n factors, and the factor (a - b) is exactly the h you need to cancel.

Extending to all real exponents

The proof above covers n = 1, 2, 3, \ldots but you also need the rule for negative exponents (like x^{-1} = 1/x), fractional exponents (like x^{1/2} = \sqrt{x}), and irrational exponents (like x^{\pi}). The full proof for arbitrary real n uses logarithmic differentiation, which depends on the derivative of \ln x — derived later in this article. Here is the argument.

Write y = x^n. Take logarithms: \ln y = n \ln x. Differentiate both sides with respect to x:

Multiply both sides by y = x^n:

The same formula. This argument works for every real n — integer, rational, or irrational — as long as x > 0 (so that \ln x is defined).

A quick check against the cases you already know:

• n = 2: f'(x) = 2x. Matches the result from the derivative article.
• n = -1: f(x) = x^{-1} = 1/x, so f'(x) = -x^{-2} = -1/x^2. Matches the second example from that article.
• n = 1/2: f(x) = \sqrt{x}, so f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.
• n = 0: f(x) = x^0 = 1, so f'(x) = 0 \cdot x^{-1} = 0. That is the constant rule again — the power rule absorbs it.

Derivatives of polynomials

Once you have the power rule, every polynomial is immediate. Take f(x) = 3x^4 - 5x^2 + 7x - 2. Each term is a constant times a power of x. Using the fact that the derivative of a sum is the sum of the derivatives (proved in the rules of differentiation):

Every polynomial, no matter how complicated, differentiates in one line.

The power rule also tells you something about degree: differentiation drops the degree of a polynomial by exactly one. A degree-n polynomial has a degree-(n-1) derivative. The second derivative has degree n - 2. After n differentiations, you reach a constant. After n + 1, you reach zero. A polynomial runs out of derivatives — its information is finite, and differentiation slowly extracts it.

The exponential function

The exponential e^x is a different kind of function — it grows multiplicatively, not polynomially. Its derivative is remarkable.

The proof from the definition

The key property of the exponential is that e^{x+h} = e^x \cdot e^h. Factor:

Since e^x does not involve h, it comes out of the limit:

Everything now rests on the limit \lim_{h \to 0} \frac{e^h - 1}{h}. This is a standard limit, and its value is 1. You can verify it numerically: at h = 0.1, \frac{e^{0.1} - 1}{0.1} \approx 1.0517; at h = 0.01, \frac{e^{0.01} - 1}{0.01} \approx 1.00502; at h = 0.001, \frac{e^{0.001} - 1}{0.001} \approx 1.0005. The values are heading steadily toward 1.

In fact, this limit is the defining property of e. The number e \approx 2.71828 is the unique base for which \lim_{h \to 0} \frac{a^h - 1}{h} = 1. Any other base a would give a different constant (namely \ln a), and the derivative of a^x would be a^x \ln a, not just a^x. The number e is special precisely because it makes this constant equal to 1.

So the derivative is

The exponential function is its own derivative. This is one of the most important facts in all of mathematics. It means the rate at which e^x is growing at any point is exactly equal to its current value at that point. A population that grows proportionally to its own size — bacteria doubling, money compounding — follows an exponential curve, and this derivative formula is why.

Think about what this says geometrically. At x = 0, e^0 = 1, so the slope of the tangent line is 1. At x = 1, e^1 \approx 2.718, so the slope is 2.718 — the curve is already climbing fast. At x = 5, e^5 \approx 148.4, so the slope is 148.4. The bigger the function gets, the steeper it climbs. This is the hallmark of exponential growth: the rate of growth is proportional to the current size, and when the constant of proportionality is 1, you get e^x.

Why general exponentials give a^x \ln a

For any base a > 0, write a^x = e^{x \ln a} (since a = e^{\ln a}). The derivative is

using the chain rule (the factor \ln a comes from differentiating the exponent x \ln a with respect to x). When a = e, \ln a = 1, and you recover \frac{d}{dx}e^x = e^x.

The logarithmic function

The natural logarithm \ln x is the inverse of e^x. Its derivative is clean and simple.

The proof from the definition

Use the logarithm property \ln(x+h) - \ln x = \ln\!\left(\frac{x+h}{x}\right) = \ln\!\left(1 + \frac{h}{x}\right):

Substitute u = h/x, so h = ux and as h \to 0, u \to 0:

The limit \lim_{u \to 0} \frac{\ln(1+u)}{u} is another standard limit, equal to 1. (This is equivalent to the limit used in the exponential proof: if \lim_{h \to 0} \frac{e^h - 1}{h} = 1, set e^h - 1 = u, so h = \ln(1+u) and u \to 0 as h \to 0, giving \lim_{u \to 0} \frac{u}{\ln(1+u)} = 1, hence \lim_{u \to 0} \frac{\ln(1+u)}{u} = 1.)

So

The derivative of \ln x is 1/x — a power function. This creates a bridge between logarithms and powers that runs through all of calculus. It also tells you something geometric: the slope of \ln x at x = 1 is 1, at x = 2 it is 1/2, at x = 10 it is 1/10. The logarithm flattens out as x grows, and it does so at a rate that is exactly the reciprocal of x.

Notice the contrast with the exponential. The derivative of e^x is e^x — as the function grows, so does its slope. The derivative of \ln x is 1/x — as x grows, the slope shrinks. This matches the graphs: e^x curves sharply upward; \ln x curves gently upward, flattening more and more. The two functions are inverses, and their derivatives reflect each other in a precise way: if y = \ln x, then x = e^y, and \frac{dy}{dx} = 1/x while \frac{dx}{dy} = e^y = x. The product of the two derivatives is 1, which is exactly the relationship between the slopes of inverse functions.

There is also a satisfying connection to integration hiding here. The fact that \frac{d}{dx}\ln x = \frac{1}{x} means that \ln x is an antiderivative of 1/x. This is the only power x^n where the power rule for integration (\int x^n \, dx = \frac{x^{n+1}}{n+1}) breaks down — because when n = -1, the formula gives \frac{x^0}{0}, which is undefined. The logarithm fills exactly this gap.

The general base

For \log_a x, use the change-of-base formula: \log_a x = \frac{\ln x}{\ln a}. Since \ln a is a constant:

When a = e, \ln a = 1, and you recover \frac{d}{dx}\ln x = \frac{1}{x}.

The standard derivatives table

Here are all the basic derivatives collected in one place. Every entry has been derived above.

These six formulas, combined with the rules of differentiation, are enough to differentiate every function you will meet in school-level calculus.

Computing one from start to finish

Time to apply the formulas. Two examples — one algebraic, one involving exponentials and logarithms.

Common confusions

• "The derivative of e^x is xe^{x-1}." No — that would be the power rule applied to e^x, treating e as the variable and x as the exponent. But e is a constant and x is in the exponent. The exponential rule, not the power rule, applies. The derivative of e^x is e^x.

• "The derivative of 2^x is x \cdot 2^{x-1}." Same mistake. The base is a constant, the variable is in the exponent. The correct derivative is 2^x \ln 2.

• "The derivative of \ln x is \ln(x-1)." This is pattern-matching from the power rule, and it does not work for logarithms. The derivative of \ln x is 1/x.

• "The power rule works only for integer exponents." It works for all real exponents: n can be 1/2, -3, \pi, or \sqrt{2}. The formula nx^{n-1} is valid whenever x^n is defined and differentiable.

• "I need to memorise the table." Understanding the proofs is better than memorising. If you forget the derivative of a^x, rewrite it as e^{x \ln a} and differentiate — you will recover a^x \ln a in one line. If you forget the derivative of \log_a x, rewrite it as \ln x / \ln a — one line gives 1/(x \ln a). The formulas are consequences, not independent facts.

Going deeper

The limit \lim_{h \to 0} \frac{e^h - 1}{h} = 1 — why it is true

This limit is not something you prove from scratch using algebra; it is intimately tied to how e is defined.

One standard definition of e is

From this, you can show that e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n. Setting x = h and expanding for small h (using the binomial approximation for large n):

This is the Taylor expansion of e^h. Subtracting 1 and dividing by h:

As h \to 0, the right side approaches 1. The higher-order terms vanish because each contains at least one factor of h.

The deeper point: e is defined so that this limit equals 1. Other bases give other constants. The number e is natural not because of some cosmic preference, but because it is the base that makes differentiation of exponentials as clean as possible — no extra constant in front.

The derivative of |x| using the power rule?

A subtle question: you know |x| = \sqrt{x^2} = (x^2)^{1/2}. Can you differentiate this using the power rule and the chain rule?

For x > 0, this is 1. For x < 0, this is -1. At x = 0, the expression x/|x| is undefined — the derivative does not exist, exactly as expected. The chain rule does not create a derivative where none exists; it correctly signals the failure.

This result \frac{d}{dx}|x| = \frac{x}{|x|} is sometimes written as \text{sgn}(x) (the sign function). It is +1 on the right, -1 on the left, and undefined at the origin — a concise encoding of the corner.

Why e is special — the characterisation theorem

The exponential function e^x is the unique function that satisfies f'(x) = f(x) and f(0) = 1. No other function does this. Here is why.

Suppose f'(x) = f(x) and f(0) = 1. Consider g(x) = f(x) \cdot e^{-x}. By the product rule:

So g'(x) = 0 everywhere, which means g is a constant. Since g(0) = f(0) \cdot e^0 = 1 \cdot 1 = 1, the constant is 1. So f(x) \cdot e^{-x} = 1, giving f(x) = e^x.

This characterisation is one of the reasons e^x is central to all of analysis. The differential equation f' = f is the simplest non-trivial differential equation, and its unique solution (with initial condition f(0) = 1) is e^x. Every more complex differential equation builds on this.

Where this leads next

You now have the derivatives of the basic building blocks. The next step is to learn how to differentiate combinations of these functions — sums, products, quotients, and compositions.

• Rules of Differentiation — the sum, product, and quotient rules, each proved from first principles.
• Chain Rule — how to differentiate a function inside another function, like e^{x^2} or \ln(\sin x).
• Derivative — the foundational article: what a derivative is, the limit definition, and the geometric meaning.
• Exponents and Powers — the algebraic background for the power rule.
• Logarithmic Functions — the algebraic background for the logarithmic derivative.