dy/dx for equations not solved for y, attaching a dy/dx factor to every y-term.A composite function is a function plugged into another: f(g(x)). The inner function g touches x first; the outer function f acts on whatever g produces. For (x² + 1)⁵, the inside is g(x) = x² + 1 and the outside is f(u) = u⁵.
Chain Rule: d/dx[ f(g(x)) ] = f'(g(x)) · g'(x)
In words: the derivative of the outside (inside left alone) times the derivative of the inside. The single most common derivative error on the CLEP exam is finishing the outer piece and forgetting to multiply by g'(x). End every chain-rule line by asking "times the derivative of the inside."
The most frequent pattern is the general power rule:
d/dx[ (g(x))ⁿ ] = n·(g(x))ⁿ⁻¹ · g'(x)
So d/dx[(x² + 1)⁵] = 5(x² + 1)⁴ · 2x = 10x(x² + 1)⁴. This also covers roots and reciprocals once you rewrite them as powers: √(x² + 1) = (x² + 1)^(1/2) and 1/(x² + 1)³ = (x² + 1)⁻³.
Treat a composite like an onion. Identify the last operation you would perform if you evaluated by hand — that is the outer function. Differentiate it, leaving the inside untouched, then multiply by the derivative of the inside. If the inside is itself a composite, repeat.
For y = sin(√(x² + 1)) there are three layers — sine (outer), square root (middle), x² + 1 (inner):
y' = cos(√(x² + 1)) · [ ½(x² + 1)^(−1/2) · 2x ]
= cos(√(x² + 1)) · x/√(x² + 1)
Every nested layer contributes one factor. A notation trap: sin²x means (sin x)² (outer square, inner sine), while sin(x²) is an outer sine of an inner square — they differentiate completely differently.
You will not always have a formula. A recurring CLEP item gives a table and asks for a composite's derivative at a point.
| x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
| 1 | 3 | 5 | 2 | 4 |
| 2 | 5 | −3 | 4 | 1 |
For h(x) = f(g(x)), the chain rule gives h'(1) = f'(g(1))·g'(1). Read the table: g(1) = 2, so f'(g(1)) = f'(2) = −3, and g'(1) = 4. Thus h'(1) = (−3)(4) = −12. Evaluate the inside first and use it as the input to f' — mixing up which value feeds which function is the classic table error.
The full circle x² + y² = 25 is not a function — most x-values give two y-values — so you cannot cleanly solve for a single y. Instead, treat y as an implicit function of x: assume a formula exists locally and differentiate accordingly. The consequence is the chain rule. Since y is a function of x:
d/dx[ yⁿ ] = n yⁿ⁻¹ · dy/dx
That extra dy/dx is the entire game. Every time you differentiate a term containing y, a dy/dx tags along; a pure-x term gets none. The procedure: differentiate both sides with respect to x, attach dy/dx to every y-term, then solve algebraically for dy/dx.
Differentiate x² + y² = 25:
2x + 2y·(dy/dx) = 0 ⟹ dy/dx = −x/y
One formula, both halves of the circle. The price of implicit differentiation is that the slope depends on both coordinates — you must supply a full point (x, y), not just an x.
A mixed term like 6xy is a product of x and y, so it needs the product rule. For the folium of Descartes x³ + y³ = 6xy:
3x² + 3y²·(dy/dx) = 6[ x·(dy/dx) + y ]
Collect every dy/dx on one side:
(3y² − 6x)·(dy/dx) = 6y − 3x² ⟹ dy/dx = (2y − x²)/(y² − 2x)
At (3, 3) (on the curve, since 27 + 27 = 54 = 6·9): dy/dx = (6 − 9)/(9 − 6) = −1.
For a tangent line: get dy/dx, plug in the point for a numerical slope m, then use point-slope form y − y₁ = m(x − x₁). Write dy/dx as a fraction N/D:
0): the numerator N = 0 (with D ≠ 0).D = 0 (with N ≠ 0).In each case, solve that condition together with the original curve equation, since the point must lie on the curve. For x² + y² = 25 with dy/dx = −x/y, a vertical tangent needs y = 0, giving (5, 0) and (−5, 0).
To find d²y/dx², differentiate dy/dx again — and every y still carries a dy/dx. For x² + y² = 25 with dy/dx = −x/y, the quotient rule gives:
d²y/dx² = −[ (1)(y) − x(dy/dx) ] / y²
Substitute dy/dx = −x/y, then use x² + y² = 25:
= −[ y + x²/y ] / y² = −(x² + y²)/y³ = −25/y³
The key move: after solving for dy/dx, substitute it back in so the final answer has no dy/dx left.
d/dx[(g)ⁿ] = n(g)ⁿ⁻¹·g' handles powers, roots, and reciprocals after rewriting.f', then multiply by the inside's derivative.dy/dx to every y-term; mixed xy-terms also need the product rule. Solve algebraically for dy/dx.dy/dx is 0; vertical tangent ⟺ denominator is 0 — solved together with the curve.y; substitute dy/dx back in to finish.d/dx[(x² + 1)⁴] =1. B) 8x(x² + 1)³. General power rule: 4(x² + 1)³ · 2x = 8x(x² + 1)³. Distractors: A) forgot the inside derivative 2x — the #1 chain-rule error. C) failed to drop the exponent by one. D) used x instead of 2x for the inside derivative. E) kept the coefficient as the inside derivative alone and dropped the 4. Fix: finish every general-power-rule line with "times the derivative of the base."
d/dx[sin(3x)] =2. B) 3cos(3x). d/dx[sin(g)] = cos(g)·g', with g = 3x, g' = 3. Distractors: A) dropped the inside factor 3. C) invented an extra negative (that belongs to cosine, not sine). D) put x inside instead of 3x. E) divided by 3 instead of multiplying. Fix: the inside derivative multiplies the outer result — for sin(ax) it is always ·a.
d/dx[√(x² + 9)] =3. D) x/√(x² + 9). Write (x² + 9)^(1/2): ½(x² + 9)^(−1/2)·2x = x/√(x² + 9). Distractors: A) forgot the inside 2x. B) did not subtract 1 from the exponent (kept a positive power). C) squared the inside derivative. E) kept a stray factor of ½ after the 2x already cancelled it. Fix: for √g, the derivative is g'/(2√g); here g' = 2x cancels the 2.
x² + y² = 25, then dy/dx =4. E) −x/y. Differentiate implicitly: 2x + 2y·(dy/dx) = 0, so dy/dx = −x/y. Distractors: A) sign error solving for dy/dx. B) inverted the ratio. C) treated y as a constant, dropping the dy/dx factor on y². D) both inverted and lost the sign. Fix: every y-term gets a dy/dx; isolate it, then read N/D.
d/dx[cos²x] =5. A) −2 sin x cos x. cos²x = (cos x)²; general power rule: 2(cos x)·(−sin x) = −2 sin x cos x. Distractors: B) dropped the negative from the derivative of cosine. C) never applied the power rule (kept the square). D) forgot the inside derivative −sin x and left 2 cos x. E) used −2 but dropped the sin x factor. Fix: cos²x is a composite — square outside, cosine inside; the inside derivative −sin x supplies the minus sign.
The differentiable functions f and g have the values below. If h(x) = f(g(x)), then h'(1) =
| x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
| 1 | 5 | 2 | 2 | 3 |
| 2 | 1 | 4 | 4 | 1 |
| 3 | 2 | −3 | 1 | 6 |
6. B) 12. h'(1) = f'(g(1))·g'(1). From the table g(1) = 2, so f'(g(1)) = f'(2) = 4, and g'(1) = 3; thus 4·3 = 12. Distractors: A) 6 used f'(1)·g'(1) = 2·3, evaluating f' at 1 instead of g(1) = 2. C) 8 used f'(2)·g(1) = 4·2, multiplying by the wrong table entry. D) −9 mixed rows f'(3)·g'(3) = −3·6. E) 2 reported f'(1) alone. Fix: compute the inside value first (g(1) = 2), plug it into f', then multiply by g'(1).
d/dx[tan(4x)] =7. A) 4sec²(4x). d/dx[tan(g)] = sec²(g)·g', with g = 4x, g' = 4. Distractors: B) forgot the inside factor 4. C) put x inside the secant instead of 4x. D) used the derivative of sec, not tan. E) squared the tangent instead of using sec². Fix: d/dx[tan(ax)] = a·sec²(ax) — the coefficient comes out front.
x³ + y³ = 6xy, the slope of the tangent line at (3, 3) is8. A) −1. Implicitly, 3x² + 3y²·y' = 6(xy' + y), so y' = (2y − x²)/(y² − 2x). At (3, 3): (6 − 9)/(9 − 6) = −1. Distractors: B) sign error in the final ratio. C) set the numerator to zero without cause. D) computed 2y − x² = −3 but forgot to divide by the denominator 3. E) assumed the denominator vanished, but 9 − 6 = 3 ≠ 0. Fix: on a product term like 6xy, apply the product rule; then substitute the full point into N/D.
x² + y² = 25 have a vertical tangent line?9. C) (5, 0). With dy/dx = −x/y, the slope is undefined where the denominator y = 0; on the circle that gives x = ±5, so (5, 0) qualifies. Distractors: A) (0, 5) and D) (0, −5) are where x = 0, giving slope 0 — those are horizontal tangents. B) (3, 4) and E) (4, 3) are ordinary points with finite nonzero slope. Fix: vertical tangent ⟺ denominator of dy/dx is zero; for −x/y that means y = 0.
d/dx[sin(√x)] =10. D) cos(√x)/(2√x). Outer sine, inner √x = x^(1/2): cos(√x) · ½x^(−1/2) = cos(√x)/(2√x). Distractors: A) dropped the inside derivative entirely. B) forgot the factor ½ from the square-root derivative. C) invented a negative sign. E) used 2√x as a multiplier instead of a denominator. Fix: the derivative of the inner √x is 1/(2√x), and it multiplies the outer cos(√x).
d/dx[cos(2x)] = −sin(2x). Which of the following correctly evaluates this claim?11. E) Incorrect; the derivative is −2 sin(2x). By the chain rule, d/dx[cos(2x)] = −sin(2x)·2 = −2 sin(2x); the student dropped the inside derivative 2. Distractors: A) and B) accept the answer, ignoring the missing chain factor. C) halves the inside factor instead of using it (writes ½ rather than 2). D) flips the sign for no reason. Fix: d/dx[cos(ax)] = −a sin(ax) — the inside factor a multiplies, it is never dropped or halved.
y² = x³ + 1, then dy/dx at the point (2, 3) is12. C) 2. Differentiate y² = x³ + 1: 2y·y' = 3x², so y' = 3x²/(2y). At (2, 3): 3(4)/(2·3) = 12/6 = 2. Distractors: A) 6 used 3x²/y alone, dropping the 2. B) 12 computed 3x² and stopped. D) 4 divided by y but not by 2. E) 3/2 used 3x/(2y) (mis-differentiated x³). Fix: implicitly, d/dx[y²] = 2y·y'; solve y' = 3x²/(2y) and then substitute the point.
1. B) 8x(x² + 1)³. General power rule: 4(x² + 1)³ · 2x = 8x(x² + 1)³. Distractors: A) forgot the inside derivative 2x — the #1 chain-rule error. C) failed to drop the exponent by one. D) used x instead of 2x for the inside derivative. E) kept the coefficient as the inside derivative alone and dropped the 4. Fix: finish every general-power-rule line with "times the derivative of the base."
2. B) 3cos(3x). d/dx[sin(g)] = cos(g)·g', with g = 3x, g' = 3. Distractors: A) dropped the inside factor 3. C) invented an extra negative (that belongs to cosine, not sine). D) put x inside instead of 3x. E) divided by 3 instead of multiplying. Fix: the inside derivative multiplies the outer result — for sin(ax) it is always ·a.
3. D) x/√(x² + 9). Write (x² + 9)^(1/2): ½(x² + 9)^(−1/2)·2x = x/√(x² + 9). Distractors: A) forgot the inside 2x. B) did not subtract 1 from the exponent (kept a positive power). C) squared the inside derivative. E) kept a stray factor of ½ after the 2x already cancelled it. Fix: for √g, the derivative is g'/(2√g); here g' = 2x cancels the 2.
4. E) −x/y. Differentiate implicitly: 2x + 2y·(dy/dx) = 0, so dy/dx = −x/y. Distractors: A) sign error solving for dy/dx. B) inverted the ratio. C) treated y as a constant, dropping the dy/dx factor on y². D) both inverted and lost the sign. Fix: every y-term gets a dy/dx; isolate it, then read N/D.
5. A) −2 sin x cos x. cos²x = (cos x)²; general power rule: 2(cos x)·(−sin x) = −2 sin x cos x. Distractors: B) dropped the negative from the derivative of cosine. C) never applied the power rule (kept the square). D) forgot the inside derivative −sin x and left 2 cos x. E) used −2 but dropped the sin x factor. Fix: cos²x is a composite — square outside, cosine inside; the inside derivative −sin x supplies the minus sign.
6. B) 12. h'(1) = f'(g(1))·g'(1). From the table g(1) = 2, so f'(g(1)) = f'(2) = 4, and g'(1) = 3; thus 4·3 = 12. Distractors: A) 6 used f'(1)·g'(1) = 2·3, evaluating f' at 1 instead of g(1) = 2. C) 8 used f'(2)·g(1) = 4·2, multiplying by the wrong table entry. D) −9 mixed rows f'(3)·g'(3) = −3·6. E) 2 reported f'(1) alone. Fix: compute the inside value first (g(1) = 2), plug it into f', then multiply by g'(1).
7. A) 4sec²(4x). d/dx[tan(g)] = sec²(g)·g', with g = 4x, g' = 4. Distractors: B) forgot the inside factor 4. C) put x inside the secant instead of 4x. D) used the derivative of sec, not tan. E) squared the tangent instead of using sec². Fix: d/dx[tan(ax)] = a·sec²(ax) — the coefficient comes out front.
8. A) −1. Implicitly, 3x² + 3y²·y' = 6(xy' + y), so y' = (2y − x²)/(y² − 2x). At (3, 3): (6 − 9)/(9 − 6) = −1. Distractors: B) sign error in the final ratio. C) set the numerator to zero without cause. D) computed 2y − x² = −3 but forgot to divide by the denominator 3. E) assumed the denominator vanished, but 9 − 6 = 3 ≠ 0. Fix: on a product term like 6xy, apply the product rule; then substitute the full point into N/D.
9. C) (5, 0). With dy/dx = −x/y, the slope is undefined where the denominator y = 0; on the circle that gives x = ±5, so (5, 0) qualifies. Distractors: A) (0, 5) and D) (0, −5) are where x = 0, giving slope 0 — those are horizontal tangents. B) (3, 4) and E) (4, 3) are ordinary points with finite nonzero slope. Fix: vertical tangent ⟺ denominator of dy/dx is zero; for −x/y that means y = 0.
10. D) cos(√x)/(2√x). Outer sine, inner √x = x^(1/2): cos(√x) · ½x^(−1/2) = cos(√x)/(2√x). Distractors: A) dropped the inside derivative entirely. B) forgot the factor ½ from the square-root derivative. C) invented a negative sign. E) used 2√x as a multiplier instead of a denominator. Fix: the derivative of the inner √x is 1/(2√x), and it multiplies the outer cos(√x).
11. E) Incorrect; the derivative is −2 sin(2x). By the chain rule, d/dx[cos(2x)] = −sin(2x)·2 = −2 sin(2x); the student dropped the inside derivative 2. Distractors: A) and B) accept the answer, ignoring the missing chain factor. C) halves the inside factor instead of using it (writes ½ rather than 2). D) flips the sign for no reason. Fix: d/dx[cos(ax)] = −a sin(ax) — the inside factor a multiplies, it is never dropped or halved.
12. C) 2. Differentiate y² = x³ + 1: 2y·y' = 3x², so y' = 3x²/(2y). At (2, 3): 3(4)/(2·3) = 12/6 = 2. Distractors: A) 6 used 3x²/y alone, dropping the 2. B) 12 computed 3x² and stopped. D) 4 divided by y but not by 2. E) 3/2 used 3x/(2y) (mis-differentiated x³). Fix: implicitly, d/dx[y²] = 2y·y'; solve y' = 3x²/(2y) and then substitute the point.