Find the Slope of the Tangent Line

On day one, we have learned about how to approximate the slope of a tangent line by using the slope of a secant line.

Example 1. Find the slope of \(f(x) = x^2\) at \(x = 0.5\).

The process becomes tedious as we become more greedy in terms of accuracy. The “almost once and for all” solution is going abstract.

  • Locate the point of interest \((0.5, 0.25)\).
  • Locate a nearby point \((0.5 + h, (0.5 + h)^2)\).
  • Draw a line connecting these two points. This is the secant line.
  • Find the slope of the secant line, \(\dfrac{(0.5 + h)^2 - 0.25}{(0.5 + h) - 0.5}\).
  • Make the nearby point closer to the point of interest.

\[\lim_{h\to 0}\dfrac{(0.5 + h)^2 - 0.25}{(0.5 + h) - 0.5} = \cdots = 1.\]

WELL DONE! We land on the accurate answer without repeating any calculation! Wait, what if my point of interest shited to 0.3 instead of 0.5? Here is the “once and for all” solution.

  • Locate the point of interest \((a, a^2)\).
  • Locate a nearby point \((a + h, (a + h)^2)\).
  • Draw a line connecting these two points. This is the secant line.
  • Find the slope of the secant line, \(\dfrac{(a + h)^2 - a^2}{(a + h) - a}\).
  • Make the nearby point closer to the point of interest.

\[\lim_{h\to 0}\dfrac{(a + h)^2 - a^2}{(a + h) - a} = \cdots = 2a.\]

Example 2. Find the slope of the tangent line of \(f(x) = \sqrt{x}\) at \(x = 4\).

Verify your answers graphically.

The Definition of the Derivatives

The slope of the tangent line is one example of the rate of change, which is mathematically known as the derivative.

The derivative of a function \(f(x)\) at \(x=a\) is denoted by \(f'(a)\), \[f'(a) = \lim_{h\to 0}\dfrac{f(a+h) - f(a)}{(a + h) - a}, \qquad \mbox{or} \qquad f'(a) = \lim_{x\to a}\dfrac{f(x) - f(a)}{x - a}.\]

Example 3. Let \(f(x) = \dfrac{1}{x^2}\). Find \(f'(3)\).


Example 4. Let \(f(x) = \dfrac{1}{\sqrt{2x+1}}\). Find \(f'(4)\).


Example 5. Let \(f(x) = \dfrac{1-2x}{1+2x}\). Find \(f'(0)\).


Verify your answers graphically.