Differentiation from first principles

Every curve has a slope that keeps changing as you move along it — unlike a straight line, there is no single number that describes it everywhere. Differentiation from first principles is the method that pins down the slope at one exact point: freeze that point, take a second point a tiny distance $h$ away, measure the slope of the chord joining them, and then watch what that chord-slope settles to as $h$ shrinks to zero. That settling value is the derivative, $f'(x)$ , built directly from the limit definition rather than from any memorised rule. It underlies every rate of change you will meet later — velocity, growth rates, marginal cost — and every rule (power, product, quotient) you will soon be allowed to use as a shortcut is itself proved by this same limit process.

The core insight

The derivative $f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ is the limiting slope of the chord from $x$ to $x+h$ as $h\to 0$ , giving the tangent slope and the instantaneous rate of change at $x$ .

The core idea

Take a function $y = f(x)$ and a fixed point $x$ on its graph. Step a small distance $h$ along the axis to the nearby point $x+h$ . The two points $(x, f(x))$ and $(x+h, f(x+h))$ determine a straight secant line (a chord), and the average rate of change of $f$ over that interval is simply its slope:

\text{slope of chord} = \frac{f(x+h)-f(x)}{h}

where: $f(x+h)-f(x)$ is the change in the function's value (rise) and $h$ is the change in $x$ (run). As $h$ shrinks toward zero, the second point slides back along the curve toward the first, and the chord rotates until it hugs the curve at $x$ — becoming the tangent line. The slope of that tangent is the derivative:

f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

where: $f'(x)$ , also written $\dfrac{dy}{dx}$ , is the derivative of $f$ at $x$ , and the limit is taken as $h$ tends to $0$ without ever equalling it. This one expression carries two readings used throughout the chapter. Geometrically, $f'(x)$ is the slope of the tangent to $y=f(x)$ at $x$ . As a rate of change, if $s(t)$ is distance travelled at time $t$ , then $s'(t)$ is instantaneous velocity — the limit of average velocity $\dfrac{s(t+h)-s(t)}{h}$ as the time interval $h$ shrinks to zero.

Figure

A curve with a fixed point and a second point at h away joined by a chord that rotates to become the tangent line as h shrinks toward zero

Finding $f'(x)$ "from first principles" always follows the same four-step routine: write $f(x+h)$ by substituting $(x+h)$ for $x$ ; form the difference quotient $\dfrac{f(x+h)-f(x)}{h}$ and simplify its numerator; cancel the common factor of $h$ (this is what resolves the $\tfrac{0}{0}$ form, since both the numerator and $h$ tend to zero together); then let $h\to 0$ in the simplified expression. It helps to notice why the $\tfrac{0}{0}$ form is unavoidable rather than a sign of a mistake. As $h\to 0$ , the second point $x+h$ slides back onto the first point $x$ , so the numerator $f(x+h)-f(x)$ shrinks to $0$ at the same time as the denominator $h$ shrinks to $0$ — direct substitution can never work here. The whole point of the algebra in Steps 1 and 2 is to rewrite the quotient in a form where the troublesome factor of $h$ is visibly cancelled before the limit is taken, leaving an expression that behaves well at $h=0$ . This is exactly the same idea used earlier in the chapter when evaluating other $\tfrac{0}{0}$ limits by factorising or rationalising first; differentiation from first principles is simply that technique applied to one very specific, recurring limit.

Derivation ADerivative of

x^2 + 3x

from the limit definition

Starting from the general definition $f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ , derive the derivative of the polynomial $f(x) = x^2 + 3x$ by expanding, simplifying, cancelling $h$ , and taking the limit, ending with the closed-form result $f'(x) = 2x + 3$ .

Step 1. Write $f(x+h)$ by replacing every $x$ in $f(x)=x^2+3x$ with $(x+h)$ :

f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h

Step 2. Form the difference $f(x+h) - f(x)$ and simplify — the $x^2$ and $3x$ terms cancel between the two brackets:

f(x+h)-f(x) = \left(x^2+2xh+h^2+3x+3h\right) - \left(x^2+3x\right) = 2xh+h^2+3h

Step 3. Divide by $h$ to form the difference quotient, then factor $h$ out of the numerator and cancel it, valid because $h\neq 0$ while the limit is being approached:

\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2+3h}{h} = \frac{h(2x+h+3)}{h} = 2x+h+3

Step 4. Let $h\to 0$ in the simplified expression, which is now a plain polynomial in $h$ with no division left, so the limit is found by direct substitution:

f'(x) = \lim_{h\to 0}(2x+h+3) = 2x+3

The result $f'(x) = 2x+3$ matches what the power and sum shortcut rules give directly ( $\frac{d}{dx}x^2=2x$ , $\frac{d}{dx}(3x)=3$ ), confirming that first-principles differentiation and the later shortcut rules always agree.

Worked example

Find the derivative of $f(x) = \sin x$ from first principles.

Step 1. Form the difference quotient directly from the definition:

\frac{f(x+h)-f(x)}{h} = \frac{\sin(x+h)-\sin x}{h}

Step 2. Simplify the numerator using the sum-to-product identity $\sin A - \sin B = 2\cos\!\left(\dfrac{A+B}{2}\right)\sin\!\left(\dfrac{A-B}{2}\right)$ with $A = x+h$ , $B = x$ :

\sin(x+h)-\sin x = 2\cos\!\left(x+\frac{h}{2}\right)\sin\!\left(\frac{h}{2}\right)

So the quotient becomes, after dividing numerator and denominator by $2$ to match the standard limit shape:

\frac{f(x+h)-f(x)}{h} = \cos\!\left(x+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2}

Step 3. Let $h\to 0$ , using the standard trigonometric limit $\lim_{h\to 0}\dfrac{\sin(h/2)}{h/2}=1$ and $\cos\!\left(x+\dfrac{h}{2}\right)\to\cos x$ :

f'(x) = \cos x \cdot 1 = \cos x

So $\dfrac{d}{dx}(\sin x) = \cos x$ , obtained purely from the limit definition, with no shortcut rule used at any step. The final answer is $\boxed{f'(x) = \cos x}$ .

Competitive stretch

The same sum-to-product route, applied to $g(x) = \cos x$ , gives $g(x+h)-g(x) = -2\sin\!\left(x+\dfrac{h}{2}\right)\sin\!\left(\dfrac{h}{2}\right)$ , and dividing by $h$ and letting $h\to 0$ produces $g'(x) = -\sin x$ . Recognising this mirror pattern — sine differentiates to cosine, cosine differentiates to negative sine — saves time in multi-part exam questions that ask for both from first principles. The minus sign is easy to lose under exam pressure, so it is worth double-checking the sum-to-product identity for cosine, $\cos A-\cos B = -2\sin\!\left(\dfrac{A+B}{2}\right)\sin\!\left(\dfrac{A-B}{2}\right)$ , term by term before dividing through by $h$ .

Common traps

Substituting $h=0$ straight into $\dfrac{f(x+h)-f(x)}{h}$ before simplifying — this always gives $\tfrac{0}{0}$ , since $f(x+h)-f(x)\to 0$ and $h\to 0$ together; the algebra must simplify the quotient first.
Expanding $(x+h)^2$ or $\sin(x+h)$ incorrectly at Step 1 — a single sign or term error there propagates through every later step, so each expansion should be checked term by term.
Treating $h$ as though it equals zero during the cancellation step — $h$ is never actually zero while the quotient is being formed; it only tends to zero in the final limit taken afterward.
Jumping straight to a memorised shortcut (writing $\frac{d}{dx}x^2=2x$ with no limit work shown) when a question explicitly asks "from first principles" — full marks require every one of the four steps to appear.

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Why do we need first principles at all once we have the power rule and other shortcuts?

The shortcut rules are themselves proved using the first-principles limit — someone has to justify why $\frac{d}{dx}x^n = nx^{n-1}$ works, and that justification is exactly this four-step process. Exam questions that say "from first principles" are testing whether you understand the definition behind the shortcuts, not just whether you can apply them, so the full limit argument must be shown even though the shortcut would give the same answer instantly.

What if the h in the denominator does not cancel out cleanly?

For every polynomial and the standard trigonometric functions on this syllabus, the numerator $f(x+h)-f(x)$ always contains a common factor of $h$ once fully expanded and simplified, precisely because $f(x+h)-f(x)\to 0$ as $h\to 0$ . If your working leaves an $h$ stuck in the denominator with no cancellation, go back and recheck the expansion in Step 1 or the algebra in Step 2 — a term has usually been dropped or mis-expanded.

Is the derivative the same thing as the tangent line itself?

No — the derivative $f'(x)$ is a single number (the slope) at a chosen point $x$ , not the whole line. The tangent line itself is built from that slope together with the point $(x, f(x))$ it passes through, using the point-slope form $y - f(x) = f'(x)(x_0 - x)$ . First-principles differentiation gives you the slope; forming the tangent line is a separate, later step.

Differentiation from first principles

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads