Calculus Reference

Tangent Lines and Linear Approximation

A tangent line uses the derivative at one point to approximate a curve nearby.

Solved Examples

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Core Formulas

Tangent Line
$$y-f(a)=f'(a)(x-a)$$
Linearization
$$L(x)=f(a)+f'(a)(x-a)$$

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of $f'(a)$ when $f'(a)\ne0$.

Equation of a Tangent Line

At a Point on a Parabola

Find the tangent line to $f(x)=x^2$ at $x=2$.

Solution Steps
  1. Find the point and derivative value at $x=2$.
  2. Use point-slope form with the point and tangent slope.
  3. State the tangent line: $y=4x-4$.
$$f(2)=4,\qquad f'(x)=2x,\qquad f'(2)=4$$$$y-4=4(x-2)$$$$y=4x-4$$

Equation of a Normal Line

Perpendicular Slope

Find the normal line to $f(x)=x^2$ at $x=2$.

Solution Steps
  1. Take the negative reciprocal of the tangent slope $4$.
  2. Use point-slope form through $(2,4)$ and state the normal line.
$$m_{\text{normal}}=-\frac14$$$$y-4=-\frac14(x-2)$$

Approximation with a Tangent Line

Approximate $\sqrt{4.1}$

Use $f(x)=\sqrt{x}$ near the convenient value $a=4$.

Solution Steps
  1. Find $f(4)$ and $f'(4)$.
  2. Build the linearization at $x=4$.
  3. Evaluate the linearization at $4.1$ and state the approximation.
$$f(4)=2,\qquad f'(x)=\frac1{2\sqrt{x}},\qquad f'(4)=\frac14$$$$L(x)=2+\frac14(x-4)$$$$\sqrt{4.1}\approx L(4.1)=2+\frac14(0.1)=2.025$$