Calculus I › Applications of derivatives › full formula sheet
The tangent line at x = a
The straight line that kisses a curve at exactly one point — and the best local copy of the function there.
Notation on this page: f′(a) is the derivative evaluated at x = a — the slope of the curve there. The point of tangency is (a, f(a)).
Before this lesson: Definition of the derivative
Where it comes from
The problem: given the curve y = f(x), find the straight line that best hugs it at the point x = a. The tempting first move is to grab a secant — a line through two points — and call it the tangent. Watch it fail on y = x² at (1, 1), using the second point (2, 4):
That is the whole idea: a tangent is a secant whose second point has slid all the way into the first. Zoom in on a smooth curve and it looks straight — the tangent is the line it looks like. Every fixed h gives a secant; only the limit h → 0 gives the tangent.
Before reading on: the secant through (a, f(a)) and (a+h, f(a+h)) has slope [f(a+h) − f(a)] / h. What single number should the tangent’s slope be — and what must be true of f for that number to exist?
Derivation
We need a line through the point (a, f(a)) whose slope hugs the curve. Two ingredients: the point-slope form of a line, and the limit definition of the derivative as the slope.
Why “provided the limit exists”? At a sharp corner like y = |x| at a = 0, secant slopes from the left approach −1 and from the right approach +1 — no single limit, no f′(0), and no tangent line of the form y = mx + b. The formula quietly assumes differentiability.
Before reading on: y = |x| at a = 0: can you write a tangent line there? What goes wrong if you plug into the formula anyway?
How to use it
The procedure, every time:
- Confirm f is differentiable at a. No corner, cusp, or vertical tangent at that x — otherwise the formula has no slope to use.
- Find the point: compute f(a). This is the y-coordinate of tangency. Sign slips here are the #1 error.
- Find the slope: compute f′(x), then evaluate at a. Differentiate first, then plug in a — never the reverse.
- Plug into y − f(a) = f′(a)(x − a). Resist the urge to rearrange until the end; the unsimplified form is easiest to check.
- Check: does your line pass through (a, f(a))? Plug x = a into your answer — you must get f(a). Ten seconds, catches most errors.
When to reach for it
Any time a problem says “tangent”, “touches”, or “linearize” at a point. It is also the first step of the linear approximation (next page) and of Newton’s method for finding roots.
The vertical-tangent trap
The formula only makes lines of the form y = mx + b. For y = ∛√x at a = 0 the tangent is the vertical line x = 0 — f′(0) is undefined, so the formula stays silent. Vertical tangents must be spotted and reported separately.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: tangent to y = x² at a = 1
- The point. f(1) = 1² = 1, so we touch at (1, 1).
- The slope. f′(x) = 2x, so f′(1) = 2.
- Plug in. y − 1 = 2(x − 1).
- Simplify. y = 2x − 2 + 1 = 2x − 1.
- Check. At x = 1: 2(1) − 1 = 1 ✓ — the line passes through (1, 1).
Your turn: Tangent to y = x² at a = 3.
Answer: y = 6x − 9.
f(3) = 9, f′(x) = 2x so f′(3) = 6; y − 9 = 6(x − 3). Check: 6(3) − 9 = 9 ✓.
Example 2 — with a root: tangent to y = √x at a = 4
- The point. f(4) = √4 = 2, so we touch at (4, 2).
- The slope. f′(x) = 1/(2√x), so f′(4) = 1/(2·2) = 1/4. (Why? √x = x1/2, derivative (1/2)x−1/2 = 1/(2√x).)
- Plug in. y − 2 = (1/4)(x − 4).
- Simplify. y = x/4 − 1 + 2 = x/4 + 1.
- Check. At x = 4: 4/4 + 1 = 2 ✓.
Your turn: Tangent to y = √x at a = 9.
Answer: y = x/6 + 3/2.
f(9) = 3, f′(9) = 1/(2·3) = 1/6; y − 3 = (1/6)(x − 9) = x/6 − 3/2 + 3. Check: 9/6 + 3/2 = 3 ✓.
Example 3 — negative a: tangent to y = x³ at a = −1
- The point. f(−1) = (−1)³ = −1, so we touch at (−1, −1).
- The slope. f′(x) = 3x², so f′(−1) = 3(1) = 3.
- Plug in — carefully. y − (−1) = 3(x − (−1)), i.e. y + 1 = 3(x + 1).
- Simplify. y = 3x + 3 − 1 = 3x + 2.
- Check. At x = −1: 3(−1) + 2 = −1 ✓.
Your turn: Tangent to y = x³ at a = 2.
Answer: y = 12x − 16.
f(2) = 8, f′(2) = 3(4) = 12; y − 8 = 12(x − 2). Check: 12(2) − 16 = 8 ✓.
Example 4 — fractions: tangent to y = 1/x at a = 2
- The point. f(2) = 1/2, so we touch at (2, 1/2).
- The slope. f′(x) = −1/x², so f′(2) = −1/4. (Why? 1/x = x−1, derivative −x−2 = −1/x².)
- Plug in. y − 1/2 = −(1/4)(x − 2).
- Simplify. y = −x/4 + 1/2 + 1/2 = −x/4 + 1.
- Check. At x = 2: −2/4 + 1 = 1/2 ✓. Sanity: 1/x is decreasing near x = 2, and the slope −1/4 is negative ✓.
Your turn: Tangent to y = 1/x at a = 1.
Answer: y = −x + 2.
f(1) = 1, f′(1) = −1; y − 1 = −(x − 1). Check: −1 + 2 = 1 ✓.
Memorization tips
- Say it aloud: “y minus f-of-a equals f-prime-of-a times x-minus-a.” The sentence rhythm matches the symbols.
- Two evaluations, then write: f at a (the point), f′ at a (the slope). Never write the formula with blanks you fill later.
- It is just point-slope form through (a, f(a)) — you already knew the shape; the only new fact is that the slope is the derivative.
- The (a, f(a)) check: plug x = a into your final line; you must recover f(a). This catches the majority of algebra slips.
- Slope sanity: glance at the graph — if the curve rises there, your slope must be positive. A sign mismatch means a wrong derivative.
- Vertical escape hatch: if f′(a) is undefined but the curve is smooth there (like ∛√x at 0), the tangent is x = a — the formula can’t say it.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the tangent line is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the tangent line formula?
The tangent line to y = f(x) at x = a is y − f(a) = f′(a)(x − a): point-slope form through (a, f(a)) with slope f′(a).
How is a tangent line different from a secant line?
A secant joins two points on the curve; the tangent is what the secant becomes when the second point slides into the first — its slope is the limit of secant slopes, f′(a).
Why must f be differentiable at a for the tangent formula?
The formula needs the slope f′(a). At a corner or cusp (like |x| at 0) the derivative doesn’t exist, so the formula gives no tangent line.
What if the tangent line is vertical?
The formula only produces lines of the form y = mx + b. A vertical tangent, like x = 0 for the cube root at 0, can’t be written that way — spot it and report it separately.
Can a tangent line touch the curve at more than one point?
Yes. The tangent to y = x³ at (1, 1) is y = 3x − 2, which meets the curve again at x = −2. Tangent means matching slope at the point, not exclusivity.
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