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Calculus Honors · AP-aligned study guide

2.1 · The derivative: difference quotients and tangent lines

Build the derivative from average rates, connect its value to a tangent slope, and decide where differentiability fails.

AP 2.1–2.4 · Reading: Larson & Edwards §2.1

Learning objectives

  • Compute a derivative from its limit definition.
  • Write and interpret a tangent-line equation.
  • Distinguish continuity from differentiability.

From a secant to a tangent

For a position function s(t)\displaystyle s(t), average velocity on [t,t+h]\displaystyle [t,t+h] is s(t+h)−s(t)h\displaystyle \frac{s(t+h)-s(t)}{h}. This is the slope of a secant line through two points on the position graph. Keep h≠0\displaystyle h\ne0 while simplifying.

The derivative is the limit of secant slopes as the second input approaches the first. If this finite limit exists, it gives instantaneous velocity or tangent slope. Its units are output units per input unit.

f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}

Evaluate the limit without dividing by zero

For f(x)=x2\displaystyle f(x)=x^2, expand (a+h)2−a2=2ah+h2\displaystyle (a+h)^2-a^2=2ah+h^2. For h≠0\displaystyle h\ne0, division by h\displaystyle h gives 2a+h\displaystyle 2a+h. Only after simplification take the limit to obtain f′(a)=2a\displaystyle f'(a)=2a.

The alternate definition lim⁡x→af(x)−f(a)x−a\displaystyle \lim_{x\to a}\frac{f(x)-f(a)}{x-a} uses the same secant geometry. Recognize either form when a limit is presented as a derivative.

Use a derivative value in context

A tangent line passes through (a,f(a))\displaystyle (a,f(a)) with slope f′(a)\displaystyle f'(a). Use point-slope form rather than substituting the slope as a height. For a height measured in meters over seconds, a derivative value of 4\displaystyle 4 means an instantaneous rise of 4\displaystyle 4 meters per second.

Differentiability implies continuity, but continuity alone is insufficient. The graph of ∣x∣\displaystyle |x| is continuous at 0\displaystyle 0, yet the left slopes approach −1\displaystyle -1 and right slopes approach 1\displaystyle 1. Corners, cusps, vertical tangents, and discontinuities require separate analysis.

y−f(a)=f′(a)(x−a)y-f(a)=f'(a)(x-a)

Worked examples

Build a derivative

Use the definition to find f′(2)\displaystyle f'(2) for f(x)=x2+1\displaystyle f(x)=x^2+1.

  1. Form (2+h)2+1−5h\displaystyle \frac{(2+h)^2+1-5}{h}.
  2. Expand and cancel h\displaystyle h for h≠0\displaystyle h\ne0: the quotient is 4+h\displaystyle 4+h.
  3. Take h→0\displaystyle h\to0 to obtain f′(2)=4\displaystyle f'(2)=4.

Write a tangent

Find the tangent to f(x)=x2\displaystyle f(x)=x^2 at x=1\displaystyle x=1.

  1. The point is (1,1)\displaystyle (1,1).
  2. The derivative definition gives f′(1)=2\displaystyle f'(1)=2.
  3. The line is y−1=2(x−1)\displaystyle y-1=2(x-1), or y=2x−1\displaystyle y=2x-1.

Common mistakes

  • Substituting zero into the difference quotient before simplifying.
  • Using a tangent slope as the point’s y-coordinate.
  • Assuming every continuous graph is differentiable.

Try it yourself

No calculator required.

1. Use the limit definition to find f′(1)\displaystyle f'(1) for f(x)=x2+3x\displaystyle f(x)=x^2+3x. Write its tangent line.

The parabola x squared plus 3x with the point (1, 4).
Use the difference quotient to find the tangent slope.
Show worked solution

The difference quotient simplifies to 5+h\displaystyle 5+h. Its limit is 5\displaystyle 5. Since f(1)=4\displaystyle f(1)=4, the tangent is y−4=5(x−1)\displaystyle y-4=5(x-1), or y=5x−1\displaystyle y=5x-1.

The parabola x squared plus 3x with the point (1, 4). The tangent at (1, 4) is y = 5x minus 1, with slope 5.
The tangent at (1, 4) is y = 5x minus 1, with slope 5.

2. Is f(x)=∣x∣\displaystyle f(x)=|x| differentiable at 0\displaystyle 0? Use one-sided slopes.

The graph of absolute x consists of two line segments meeting at the origin.
Compare the slope on each side of the origin.
Show worked solution

The left difference quotients equal −1\displaystyle -1, while the right quotients equal 1\displaystyle 1. Since their limits differ, f′(0)\displaystyle f'(0) does not exist, although f\displaystyle f is continuous.

The graph of absolute x consists of two line segments meeting at the origin. The left slope is minus 1 and the right slope is 1, so no derivative exists at zero.
The left slope is minus 1 and the right slope is 1, so no derivative exists at zero.

3. Express lim⁡h→0(3+h)2−9h\displaystyle \lim_{h\to0}\frac{(3+h)^2-9}{h} as a derivative and evaluate it.

Show worked solution

This is f′(3)\displaystyle f'(3) for f(x)=x2\displaystyle f(x)=x^2. The quotient is 6+h\displaystyle 6+h, so the limit is 6\displaystyle 6.

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