Derivatives, Limits, and Tangent Lines — Lecture Notes

Limits are the foundation of calculus; derivatives generalize the slope of a straight line to arbitrary functions.
Secant line slope as an average rate of change: for a function f and two points, the slope is
$\text{AvgSlope} = \frac{f(b)-f(a)}{b-a}$
which gives an average rate of change between a and b. Example: for (f(x)=x^2) between x=1 and x=2,
(f(2)=4, f(1)=1) → average rate of change = (\frac{4-1}{2-1}=3) units per unit.
Real-time (instantaneous) rate of change requires looking at a single point, not two.
The instantaneous rate of change at (a) is defined as the derivative:
$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$
This limit corresponds to the slope of the tangent line to the graph of (f) at (x=a).
Tangent line: the line that touches the graph at (x=a) exactly once; its slope is the derivative (f'(a)).
In many problems we work with the velocity interpretation: if (f) models distance, then (f'(a)) is the instantaneous velocity at time (a).
Velocity vs. speed:
- Velocity carries direction; speed is the absolute value of velocity.
- In problems, we often discuss velocity (signed rate) but may report speed when direction is not needed.

Primary difference quotient (two-point form):
$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$
Change of variable form (one-point, using (h) with (x=a+h)):
$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$
Relationship between the two forms: they are equivalent representations of the same limit.

Example: (f(x)=x^2). Average rate of change between 1 and 2:
- (f(2)=4, f(1)=1)
- (\text{AvgSlope}=\frac{4-1}{2-1}=3).
Instantaneous rate of change at (x=1) using the first definition:
- Compute (\frac{f(x)-f(1)}{x-1}=\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1) (for (x\neq 1)).
- Take the limit as (x\to 1): (f'(1)=2).
General linear function: (f(x)=5x+7).
- The derivative should be the slope; compute via the derivative limit:
- (f'(a)=\lim{x\to a}\frac{(5x+7)-(5a+7)}{x-a}=\lim{x\to a}\frac{5(x-a)}{x-a}=5).
- This shows that the derivative of a linear function is its constant slope, for any (a).
Special case: derivative of a constant is zero: if (f(x)=c), then (f'(a)=0).

Choose (x=a+h). Then
- Numerator: (f(a+h)-f(a)) (unchanged in form).
- Denominator: (h) (since (x-a=(a+h)-a=h)).
The derivative becomes
$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$
This form is advantageous because it simplifies many cancellations and is easier to manipulate for many functions.

Let (a=1). Then
$f'(1)=\lim<em>{h\to 0}\frac{(1+h)^2-1}{h}=\lim</em>{h\to 0}\frac{1+2h+h^2-1}{h}=\lim<em>{h\to 0}\frac{2h+h^2}{h}=\lim</em>{h\to 0}(2+h)=2.$
This matches the previous result, illustrating consistency between the two forms.

For (f(x)=mx+b), compute using the limit:
- (f'(a)=\lim{x\to a}\frac{(mx+b)-(ma+b)}{x-a}=\lim{x\to a}\frac{m(x-a)}{x-a}=m).
This confirms that the derivative of a line is its slope, and is independent of the point (a).
Special case: derivative of a constant (i.e., (m=0)) is zero, as expected.

For higher powers or more complex expressions (e.g., (x^3-1)), the difference quotient can become unwieldy to simplify by factoring.
Factoring general polynomials like (x^3-1=(x-1)(x^2+x+1)) helps, but becomes tedious for larger powers.
The (h)-form often provides the more practical path to simplification and computation for many cases.

Example: Find (f'(16)) for (f(x)=\sqrt{x}) using the (h)-form with a = 16:
- (f'(16)=\lim_{h\to 0}\frac{\sqrt{16+h}-4}{h}).
- Multiply numerator and denominator by the conjugate:
  $\frac{\sqrt{16+h}-4}{h}\cdot \frac{\sqrt{16+h}+4}{\sqrt{16+h}+4}=\frac{(16+h)-16}{h(\sqrt{16+h}+4)}=\frac{h}{h(\sqrt{16+h}+4)}=\frac{1}{\sqrt{16+h}+4}.$
- As (h\to 0), this tends to (\frac{1}{4+4}=\frac{1}{8}).
Therefore, (f'(16)=\dfrac{1}{8}).

Tangent line at (x=a) has slope (f'(a)) and passes through ((a,f(a))).
Point-slope form of the tangent line:
$y-f(a)=f'(a)\bigl(x-a\bigr).$
Alternatively, slope-intercept form: if desired, convert to (y=mx+b) with (m=f'(a)) and (b=f(a)-a f'(a)).
Example: For (f(x)=\sqrt{x}) at (x=16):
- (f(16)=4), (f'(16)=\dfrac{1}{8}).
- Tangent line: $y-4=\frac{1}{8}(x-16)$ or equivalently $y=\frac{1}{8}x+2.$
Application: Linear approximation. For (x) near 16, (\sqrt{x}) is well approximated by the tangent line:
- (\sqrt{17}\approx \frac{1}{8}\cdot 17+2=4.125) while true value is (\sqrt{17}\approx 4.1231).
- Accuracy improves as you stay closer to the point of tangency (here, 16).
Intuition: Zooming in on a curve makes it look like its tangent line; this justifies using the line for local approximations.

Instantaneous rate of change (derivative) is the core concept behind velocity in physics as well as rate-of-change in many other fields (economics, biology, engineering).
The derivative lets you relate small changes in input to changes in output, enabling linear approximations (first-order Taylor approximation) near a point.
Distinguishing speed vs. velocity is important in problems where direction matters (velocity) vs. magnitude only (speed).

You may see problems that require using the limit definition of the derivative, possibly with non-polynomial functions (e.g., roots, higher powers, rational expressions).
Often, exams focus on one of three common challenging forms:
- Square roots (use conjugates to simplify).
- Higher powers (expand or use algebraic tricks).
- Fractional expressions (simplify carefully or use the h-substitution).
There is typically at least one problem where you must identify the function and the point from a given limit expression (a multiple-choice derivative definition problem).

Problem: Find the tangent line for (f(x)=3x^2) at (x=1).
- Evaluate: (f(1)=3).
- Derivative: (f'(x)=6x) so (f'(1)=6).
- Tangent line in point-slope form:
  $y-f(1)=f'(1)(x-1)\;\,\Rightarrow\ y-3=6(x-1).$
- Convert to slope-intercept form: (y=6x-3).
Quick review: At a point (a), the tangent line has slope (f'(a)) and passes through ((a,f(a))). The line equation is either
$y-f(a)=f'(a)(x-a)$ or, equivalently, (y= f'(a)x + [f(a)-a f'(a)].)

Practice both forms of the derivative to become fluent: the two-point quotient and the h-substitution quotient.
When facing a square root or higher-power expression, try the conjugate or the h-substitution first to simplify the limit.
Always verify special cases: derivative of a constant is 0; derivative of a linear function is its slope.
Use tangent-line reasoning to motivate linear approximations and provide intuition for why derivatives are useful.
For quiz and exam problems, read limits carefully and identify whether you should factor, rationalize, or shift variables to an h-substitution for cancellation.