Derivatives: definition, rules, and tangent lines

2026-10-22 huynhvietkhanh Comments: 0

Calculate derivatives from the definition and differentiation rules, distinguish a derivative value from a derivative function, and write a tangent line.

High schoolFunctions, Graphs & Tangents / Sequences, Limits & Derivatives

Content updated: October 22, 2026

The derivative at a point is the limit of the difference quotient as two points approach one another. Differentiation rules avoid repeating the limit calculation, but apply only where the function and the expressions involved satisfy the required conditions.

Objectives: calculate derivatives from the definition and rules; distinguish a derivative at a point from the derivative function; write a tangent line when the point of tangency is known.

Prerequisites: finite limits, algebraic identities, and trigonometric functions measured in radians.

On this page: Definition · Method · Examples · Practice · Further learning

1. The derivative at a point

Let $f$ be defined on an open interval containing $x_0$. If the following limit exists and is finite, it is called the derivative of $f$ at $x_0$:

$$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}h.$$

The quotients are considered only for $x\ne x_0$ or $h\ne0$. Direct substitution leading to $0/0$ does not establish a conclusion; simplify or estimate before taking the limit. If the derivative exists at every point of an interval, these values define the derivative function $f'$ on that interval.

A finite derivative implies continuity: for $h\ne0$, the difference $f(x_0+h)-f(x_0)$ equals $h$ times the quotient above, so it tends to 0. The converse fails: $|x|$ is continuous at 0, but the quotient $|h|/h$ has right-hand limit 1 and left-hand limit $-1$, so the derivative does not exist at 0.

2. Formulas and rules with conditions

For a constant $C$ and an integer $n\ge1$, we have

$$C'=0,\quad (x)'=1,\quad (x^n)'=nx^{n-1},\quad (1/x)'=-1/x^2\ \ (x\ne0).$$

For $x>0$, $(\sqrt x)'=1/(2\sqrt x)$. This formula does not yield a finite derivative at 0. When angles are measured in radians, $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$. Also,

$$(\tan x)'=\frac1{\cos^2x}=1+\tan^2x\quad(\cos x\ne0),$$

$$(\cot x)'=-\frac1{\sin^2x}=-(1+\cot^2x)\quad(\sin x\ne0).$$

These are the basic formulas used in this lesson; the first two examples below calculate directly from the definition. For $u,v$ differentiable at the point under consideration, we have

$$(u\pm v)'=u'\pm v',\quad (Cu)'=Cu',\quad (uv)'=u'v+uv'.$$

$$(1/v)'=-\frac{v'}{v^2},\qquad (u/v)'=\frac{u'v-uv'}{v^2}\quad(v\ne0).$$

At a point with a finite derivative, the tangent at $M_0=(x_0;f(x_0))$ has slope $f'(x_0)$ and equation

$$y=f'(x_0)(x-x_0)+f(x_0).$$

Use both the function value and derivative at the same input $x_0$. A derivative of 0 gives a horizontal tangent; the formula above does not describe a vertical tangent through a finite derivative.

3. Six worked examples

Example 1. A derivative value from the definition

Let $f(x)=x^2+1$ and calculate $f'(2)$. Since $f(2)=5$, for $x\ne2$ we have

$$\frac{f(x)-f(2)}{x-2}=\frac{x^2-4}{x-2}=x+2.$$

Taking the limit as $x\to2$ gives $f'(2)=4$. The factor is canceled at nearby inputs different from 2, as required by the domain of the quotient.

Example 2. A derivative function from the definition

For $f(x)=x^3$ and $x\ne x_0$, the difference-of-cubes identity makes the difference quotient equal to $x^2+xx_0+x_0^2$. As $x\to x_0$, it tends to $3x_0^2$. Since $x_0$ is arbitrary in $\mathbb R$, it follows that $f'(x)=3x^2$ on the whole real line.

Example 3. Differentiating a polynomial

For $f(x)=x^3-x^2+x-1$, the sum and difference rules imply $f'(x)=3x^2-2x+1$. Every term is differentiable on $\mathbb R$, so the formula holds for every real input.

Example 4. Constant coefficients on individual terms

For $f(x)=2x^3-5x^2+3x+2$, we have

$$\begin{aligned}f'(x)&=2(3x^2)-5(2x)+3,\\&=6x^2-10x+3.\end{aligned}$$

The final constant has derivative $0$; retain each other coefficient before multiplying it by the derivative of the corresponding term.

Example 5. The quotient rule

Let $f(x)=(x^2+1)/(x^3-2x+1)$ on the domain $x^3-2x+1\ne0$. Applying the quotient rule,

$$\begin{aligned}f'(x)&=\frac{2x(x^3-2x+1)-(3x^2-2)(x^2+1)}{(x^3-2x+1)^2},\\&=\frac{-x^4-5x^2+2x+2}{(x^3-2x+1)^2}.\end{aligned}$$

Expand the numerator only after writing the correct order $u'v-uv'$. Even if cancellation is possible at another step, points where the original denominator is 0 remain outside the function's domain.

Example 6. A tangent to the sine graph

Let $f(x)=\sin x$ and $x_0=\pi/6$. Then $f(x_0)=1/2$ and $f'(x_0)=\cos(\pi/6)=\sqrt3/2$. Thus the tangent is

$$y=\frac{\sqrt3}{2}\left(x-\frac\pi6\right)+\frac12=\frac{\sqrt3}{2}x+\frac12-\frac{\sqrt3\pi}{12}.$$

Substituting $x=\pi/6$ gives $y=1/2$, the correct point of tangency; the slope is the derivative value already calculated.

4. Practice

Differentiate $g(x)=x^4-2x^2+7$.
Write the tangent to $y=x^2+1$ at the point with horizontal coordinate 2.
For $f(x)=\sqrt x$ on $[0;+\infty)$, examine the right-hand difference quotient at 0. Is there a finite derivative there?

Hints and answers

$g'(x)=4x^3-4x$ on $\mathbb R$.
By Example 1, the point of tangency is $(2;5)$ and the slope is 4. The equation is $y=4(x-2)+5=4x-3$.
For $h>0$, the quotient is $\sqrt h/h=1/\sqrt h\to+\infty$. There is no finite right-hand derivative; the two-sided definition on an open interval also does not apply at the domain endpoint 0.

5. Common mistakes

Distinguish the derivative function from the derivative at a point. An infinite limit is not a finite derivative under this lesson's definition. In the product and quotient rules, retain both terms and the denominator condition. For a tangent line, do not omit $f(x_0)$ or use a value at a different point.

6. Further learning

Derivatives of basic functions (English lesson) · The chain rule (English lesson) · The tangent line at a point (English lesson) · Derivatives of exponential and logarithmic functions (English lesson).

Original exercise material on Scribd.