Lesson 8: Implicit Differentiation and Derivatives of Inverse FunctionsAP

Some curves are not functions y(x) written out in the open — they are *relations*, like the circle x^2 + y^2 = 25, where y is tangled up with x. Implicit differentiation (CED topics 3.2–3.5) differentiates such relations directly, treating y as an (unknown) function of x and letting the chain rule do the work. The same lesson covers the derivative of an inverse function, another chain-rule consequence and one of the exam's most predictable table problems.

The method

Differentiate both sides with respect to x. Every time you differentiate a term containing y, the chain rule attaches a factor of dy/dx (often written y'):

• d/dx (y^2) = 2y·y'
• d/dx (xy) = y + x·y' (product rule, then chain rule on the y factor)
• d/dx (sin y) = cos y · y'

Then collect every y' term on one side and solve. The result typically involves both x and y — that is expected: the slope of an implicitly defined curve genuinely depends on where on the curve you stand.

Worked Example 1: the circle, and its tangent lines

Problem. For x^2 + y^2 = 25, find dy/dx and the tangent lines at (3, 4) and (3, −4).

Differentiate:  2x + 2y·y' = 0  →  y' = −x/y
At (3, 4):  y' = −3/4;  tangent:  y − 4 = −(3/4)(x − 3)
At (3, −4): y' = 3/4;   tangent:  y + 4 = (3/4)(x − 3)

Answer: dy/dx = −x/y, symmetric slopes at the mirror points — geometrically obvious once seen: radii to (3, ±4) have slopes ±4/3, and tangents are perpendicular to radii. Trying to solve for y first (y = ±sqrt(25 − x^2)) gives two half-circles and twice the work; implicit differentiation handles the whole curve at once. Note y' is undefined at (0, ±5)... it is undefined where y = 0, namely (±5, 0) — exactly where the circle has *vertical* tangents. The algebra and geometry agree.

Worked Example 2: a mixed curve, and a second derivative

Problem. The curve x^2 + xy + y^2 = 7 passes through (1, 2). Find the slope there, then find d^2y/dx^2 for the circle of Example 1 at (3, 4).

Differentiate x^2 + xy + y^2 = 7:
  2x + (y + x·y') + 2y·y' = 0
  y'(x + 2y) = −(2x + y)   →   y' = −(2x + y)/(x + 2y)
At (1, 2):  y' = −(2 + 2)/(1 + 4) = −4/5

Second derivative for the circle, from y' = −x/y:
  y'' = −[ (1)(y) − x·y' ] / y^2  = −(y − x·y')/y^2
  Substitute y' = −x/y:  y − x(−x/y) = y + x^2/y = (x^2 + y^2)/y
  y'' = −(x^2 + y^2)/y^3 = −25/y^3   (using x^2 + y^2 = 25)
  At (3, 4):  y'' = −25/64

Answers: slope −4/5 at (1, 2); y''(3) = −25/64 for the circle (concave down on the upper half — as the picture demands). Second-derivative implicit problems follow a fixed script: differentiate the y' relation again, then *substitute the first derivative and the original equation* to clean up. The substitution step is where the marks are.

Derivatives of inverse functions

If g is the inverse of a differentiable function f, and f'(g(b)) ≠ 0, then:

Geometrically this is reflection in the line y = x: reflecting swaps rise and run, so slopes reciprocate. The same idea yields d/dx (arctan x) = 1/(1 + x^2), d/dx (arcsin x) = 1/sqrt(1 − x^2), and by the chain rule d/dx arctan(2x) = 2/(1 + 4x^2).

Worked Example 3: inverse derivative from a table (graphical/numerical)

Problem. f is differentiable and increasing, with f(1) = 2, f(2) = 5, f'(1) = 3, f'(2) = 4. Let g be the inverse of f. Find g'(5) and g'(2).

g'(5) = 1/f'(g(5)).  g(5) is the x with f(x) = 5, namely x = 2.
  g'(5) = 1/f'(2) = 1/4
g'(2) = 1/f'(g(2)) = 1/f'(1) = 1/3

Answers: g'(5) = 1/4, g'(2) = 1/3. The two-step discipline — *first* find which input produces the given output, *then* take the reciprocal of f' at that input — defeats the standard trap of computing 1/f'(5) from values that do not exist.

Real-world application: sensitivity in both directions

An exchange bureau converts dollars to euros by E = f(D) with f'(D) ≈ 0.85 euros per dollar near today's volume: one extra dollar buys about 0.85 extra euros. A traveller asks the reverse question — how many extra dollars must I bring per extra euro I want? That is the derivative of the inverse: dD/dE = 1/(dE/dD) ≈ 1/0.85 ≈ 1.18 dollars per euro. Every "how much input per unit output" question in engineering and economics is an inverse-derivative question: dose per unit of drug response, energy per unit of temperature rise, study hours per point of score. The reciprocal structure is why small forward sensitivities become large reverse ones — when f'(a) is near 0, the inverse is extremely sensitive at b, and the formula shows it before any numbers are computed.

Common mistakes

• Differentiating y-terms as if y were x: writing d/dx (y^2) = 2y instead of 2y·y'. Every y-term owes a y' factor to the chain rule.
• Product-rule amnesia on xy terms: d/dx (xy) is y + x·y', two terms, not x·y'.
• In inverse problems, evaluating f' at b instead of at g(b). g'(7) needs f' at the input whose output is 7 — find that input first.
• Sign errors when isolating y': move all y' terms to one side *before* dividing; trying to divide term-by-term mid-collection is where negatives get lost.

AP Exam Experience

1. (No calculator.) If x^2 + y^2 = 13, then dy/dx at the point (2, 3) is

• (A) −2/3
• (B) 2/3
• (C) −3/2
• (D) 3/2

Answer: (A). 2x + 2y·y' = 0 gives y' = −x/y = −2/3 at (2, 3). (C) is the reciprocal-slip: −y/x. Tangents to a circle at (2, 3) have slope −2/3 because the radius has slope 3/2 and the tangent is perpendicular.

2. Let f be differentiable with f(4) = 7 and f'(4) = 2. If g is the inverse of f, then g'(7) =

• (A) 2
• (B) 1/2
• (C) −1/2
• (D) 1/7

Answer: (B). g'(7) = 1/f'(g(7)) and g(7) = 4 (since f(4) = 7), so g'(7) = 1/f'(4) = 1/2. (D) is the tempting nonsense answer 1/7; the reciprocal acts on the *derivative*, not on the input.

3. (No calculator.) d/dx arctan(x^2) =

• (A) 2x/(1 + x^4)
• (B) 1/(1 + x^4)
• (C) 2x/sqrt(1 − x^4)
• (D) −2x/(1 + x^4)

Answer: (A). Chain rule with d/du arctan u = 1/(1 + u^2): with u = x^2, (1/(1 + (x^2)^2))·2x = 2x/(1 + x^4). (C) is the arcsin pattern pasted onto arctan.

Free-response question. Consider the curve xy + y^2 = 6.

(a) Show that dy/dx = −y/(x + 2y). (b) Write the equation of the tangent line at the point (1, 2). (c) Find d^2y/dx^2 at (1, 2) and state whether the curve is concave up or concave down there.

Scoring guideline.

• (a) 2 points. 1 point for differentiating both sides correctly: y + x·y' + 2y·y' = 0 (product rule on xy, chain rule on y^2). 1 point for collecting y' terms, y'(x + 2y) = −y, and solving to the stated form. A derivation with a missing y' factor earns neither point.
• (b) 1 point. Slope at (1, 2): −2/(1 + 4) = −2/5; tangent y − 2 = −(2/5)(x − 1).
• (c) 2 points. 1 point for a correct second-differentiation setup, e.g. differentiating y + (x + 2y)y' = 0 to get y' + (1 + 2y')y' + (x + 2y)y'' = 0, i.e. y'' = −2y'(1 + y')/(x + 2y). 1 point for substitution of y' = −2/5 at (1, 2): y'' = −2(−2/5)(3/5)/5 = 12/125, with the conclusion concave up (since y'' > 0).

Additional worked practice

Problem. The curve x + y = cos y passes through (1, 0). Find dy/dx there and the tangent line.

Differentiate implicitly:  1 + y' = −sin y · y'
At (1, 0): sin 0 = 0, so 1 + y' = 0 → y' = −1.
Tangent: y − 0 = −1(x − 1), i.e. y = 1 − x.

Answer: dy/dx = −1 at (1, 0). Two quiet lessons hide in this short computation. First, verify the point before differentiating: 1 + 0 = 1 = cos 0 ✓ — implicit problems occasionally plant points that are not on the curve at all. Second, the chain-rule factor y' multiplies sin y on the right, and it happened to vanish at this point (sin 0 = 0), collapsing the equation to something trivial; do not let an easy instance convince you the factor is optional — at a generic point of this curve, y' = −1/(1 + sin y), a formula in which the factor did all the work.

Extension: normal lines, horizontal tangents, and a harder inverse

The normal line to a curve at a point is perpendicular to the tangent there, so its slope is the negative reciprocal of dy/dx. For the circle x^2 + y^2 = 25 at (3, 4), the tangent slope is −3/4, so the normal has slope 4/3: y − 4 = (4/3)(x − 3). (Reassuringly, this is the line through the origin and (3, 4) — the radius — as circle geometry demands.)

Horizontal tangents on an implicit curve occur where dy/dx = 0, i.e. where the *numerator* of the derivative vanishes (and the denominator does not). For x^2 + xy + y^2 = 7, dy/dx = −(2x + y)/(x + 2y) = 0 requires y = −2x. Substituting into the curve: x^2 − 2x^2 + 4x^2 = 3x^2 = 7, so x = ±sqrt(7/3) ≈ ±1.53, giving the points (sqrt(7/3), −2 sqrt(7/3)) and its negative — the topmost and bottommost points of this tilted ellipse, found without ever solving for y.

A harder inverse computation: f(x) = x^5 + x + 1 is strictly increasing (f'(x) = 5x^4 + 1 > 0), hence invertible. Its inverse g satisfies g(3) = 1 because f(1) = 3, so g'(3) = 1/f'(1) = 1/(5 + 1) = 1/6. Note that nobody can write g explicitly — x^5 + x + 1 = y has no usable closed-form solution — yet its derivative at a point is a two-line computation. That asymmetry is exactly why the exam loves inverse derivatives.