Difficulty: Introductory–Intermediate | Prerequisites: Limits (Ch. 2.1), continuity, algebra of radicals and polynomials
The derivative of a function gives you the slope of the tangent line at any point. You can find it the hard way (limit definition) or the fast way (power rule and friends). Once you have the derivative, setting it equal to zero tells you where the original function has horizontal tangent lines, which turns up constantly on exams.
Derivative (of f at x)
The instantaneous rate of change of f at the point x, written f'(x) or df/dx. Formally, it is the limit of the difference quotient as h approaches 0. In simple terms, it tells you the slope of the curve at a single point.
Limit definition of the derivative
f'(x) = lim (h→0) [f(x+h) − f(x)] / h. This is the "from scratch" method: you build the difference quotient, simplify until the h in the denominator cancels, then take the limit. Think of it as zooming in on the curve until the secant line becomes the tangent line.
Power rule
If f(x) = xⁿ, then f'(x) = n·xⁿ⁻¹. This is the shortcut that replaces the limit definition for any power function. In simple terms, bring the exponent down as a coefficient, then subtract one from the exponent.
Derivative function (f')
A new function that gives the slope of f at every point in its domain. Where f is increasing, f' is positive; where f is decreasing, f' is negative; where f has a peak or valley, f' is zero.
Horizontal tangent line
A tangent line with slope zero. You find them by solving f'(x) = 0. Think of it as the points where the curve momentarily levels off.
Vertical tangent line
A tangent line with undefined (infinite) slope. At these points, f' does not exist. For f(x) = 2√x, this happens at x = 0 because f'(x) = 1/√x, which blows up as x→0.
Tangent line equation
The line that just touches the graph of f at a point (a, f(a)). Its equation is y − f(a) = f'(a)(x − a). You need two things: the y-value f(a) and the slope f'(a).
Start with the definition: f'(x) = lim (h→0) [f(x+h) − f(x)] / h
Substitute your function into f(x+h) and f(x)
Simplify the numerator until you can cancel the h in the denominator
Take the limit as h→0
Step 1: Write the difference quotient
f'(x) = lim (h→0) [2√(x+h) − 2√x] / h
Step 2: Rationalise the numerator
Multiply top and bottom by the conjugate: 2√(x+h) + 2√x.
The numerator becomes:
[2√(x+h) − 2√x][2√(x+h) + 2√x] = 4(x+h) − 4x = 4h
So the expression simplifies to:
4h / [h(2√(x+h) + 2√x)] = 4 / [2√(x+h) + 2√x]
Step 3: Take the limit
As h→0, the expression becomes:
4 / [2√x + 2√x] = 4 / (4√x) = 1/√x
So f'(x) = 1/√x.
Whenever your function contains a square root, the conjugate trick is almost always the move. Multiply top and bottom by (√(something) + √(something else)) to clear the radical from the numerator. This creates a difference of squares that cancels the h.
For f(x) = 2√x = 2x^(1/2), apply the power rule:
f'(x) = 2 · (1/2) · x^(1/2 − 1) = x^(−1/2) = 1/√x
This matches the limit-definition result. The power rule works for fractional exponents too, not just whole numbers.
To write the equation of the tangent line to f(x) = 2√x at the point (9, f(9)):
Compute f(9) = 2√9 = 6. The point is (9, 6).
Compute the slope: f'(9) = 1/√9 = 1/3.
Plug into point-slope form: y − 6 = (1/3)(x − 9).
A horizontal tangent means slope = 0, so set f'(x) = 0 and solve.
Worked example: g(x) = x³ − 3x² − 45x + 19
Differentiate: g'(x) = 3x² − 6x − 45.
Set g'(x) = 0: 3x² − 6x − 45 = 0.
Divide by 3: x² − 2x − 15 = 0.
Factor or use the quadratic formula: x = [2 ± √(4 + 60)] / 2 = [2 ± 8] / 2.
Solutions: x = 5 and x = −3.
So g has horizontal tangent lines at x = 5 and x = −3.
For f(x) = 2√x, f'(x) = 1/√x. At x = 0, the denominator is zero, so f'(0) does not exist. The graph has a vertical tangent there: the curve rises infinitely steeply.
All tangent lines to f(x) = 2√x for x > 0 have positive slopes (since 1/√x > 0 for all positive x). The function is always increasing on its domain.
Limit definition of the derivative
f'(x) = lim (h→0) [f(x+h) − f(x)] / h
Power rule
d/dx [xⁿ] = n·xⁿ⁻¹ (works for all real n, including fractions and negatives)
Constant multiple rule
d/dx [c·f(x)] = c·f'(x)
Sum/difference rule
d/dx [f(x) ± g(x)] = f'(x) ± g'(x)
Point-slope form of a tangent line
y − f(a) = f'(a)(x − a)
Students often forget to rationalise the numerator when using the limit definition on square-root functions. Without the conjugate trick, the h in the denominator never cancels and you are stuck.
Applying the power rule to 2√x without rewriting it first. You must write 2√x as 2x^(1/2) before differentiating. The power rule applies to xⁿ, not to √x directly.
Confusing "horizontal tangent" with "vertical tangent." Horizontal tangent means f'(x) = 0 (slope is zero). Vertical tangent means f'(x) does not exist (slope is infinite).
Forgetting that the power rule applies to fractional and negative exponents, not just positive integers. If the exponent is 1/2, you still bring it down and subtract 1.
⚠️ The limit definition of the derivative is almost certain to appear on Exam 1. You will be asked to compute a derivative "from the definition" or "using the limit definition" for a specific function. Practise the conjugate trick for roots and the algebra of expanding polynomials.
⚠️ Finding tangent lines is a two-step process that exams love: first get the slope via f'(a), then plug into point-slope form. Do not forget to evaluate f(a) for the y-coordinate.
⚠️ "Find where the function has a horizontal tangent line" = set f'(x) = 0 and solve. This is one of the most common Exam 1 question types.
⚠️ Know the difference between where f'(x) = 0 (horizontal tangent) and where f'(x) does not exist (vertical tangent or corner). Both may be asked about.
True or False: The power rule can only be used when the exponent is a positive integer.
Answer: False. It works for any real exponent, including fractions and negatives.
Fill in the blank: To find where a function has a horizontal tangent line, set ______ equal to zero.
Answer: f'(x) (the derivative).
True or False: If f'(a) does not exist, the function f definitely has a vertical tangent at x = a.
Answer: False. The derivative can fail to exist for other reasons too, such as a corner or a cusp.
Fill in the blank: The tangent line to f(x) = 2√x at x = 9 has slope ______.
Answer: 1/3.
True or False: When using the limit definition on a function with a square root, the standard technique is to multiply by the conjugate.
Answer: True.
Q: Use the limit definition to find the derivative of f(x) = 3x² + 1.
A: f'(x) = lim (h→0) [3(x+h)² + 1 − (3x² + 1)] / h = lim (h→0) [6xh + 3h²] / h = lim (h→0) (6x + 3h) = 6x.
Q: Find f'(x) for f(x) = 5x^(3/2) using the power rule.
A: f'(x) = 5 · (3/2) · x^(1/2) = (15/2)√x.
Q: Find the equation of the tangent line to g(x) = x³ − 3x² − 45x + 19 at x = 5.
A: g(5) = 125 − 75 − 225 + 19 = −156. g'(x) = 3x² − 6x − 45, so g'(5) = 75 − 30 − 45 = 0. The tangent line is y = −156 (a horizontal line).
Q: For f(x) = 2√x, explain why f'(0) does not exist and what this means geometrically.
A: f'(x) = 1/√x, which is undefined at x = 0 (division by zero). Geometrically, the graph of f has a vertical tangent line at the origin: the curve rises infinitely steeply there.
Q: Find all x-values where h(x) = x⁴ − 8x² has a horizontal tangent line.
A: h'(x) = 4x³ − 16x = 4x(x² − 4) = 4x(x − 2)(x + 2). Setting h'(x) = 0 gives x = 0, x = 2, and x = −2.
The limit definition of the derivative builds directly on the limit techniques from Chapter 2.1. If you are shaky on limits (especially limits involving indeterminate forms like 0/0), review those first, because the limit definition will always produce a 0/0 form that you need to resolve.
Horizontal tangent lines connect to optimisation later in the course: the critical points where f'(x) = 0 are exactly where maxima and minima can occur. This idea comes back in a big way when you study curve sketching and applied optimisation problems.
The tangent line itself is the foundation of linear approximation (Section 3.9 in most Calculus I textbooks). The tangent line is the best linear approximation to f near the point of tangency.
derivative, limit definition of derivative, difference quotient, power rule, tangent line, point-slope form, horizontal tangent, vertical tangent, slope of a curve, instantaneous rate of change, f'(x), dy/dx, differentiation, rationalising the numerator, conjugate, MTH 132, Calculus I, MSU, Michigan State