eˣ, ln x, aˣ, logₐ x — and their chain-rule composites e^(g(x)) and ln(g(x)).(f⁻¹)'(b) = 1/f'(f⁻¹(b)), including from a table of values.arcsin, arccos, arctan.eˣ.Four results to know cold:
d/dx[ eˣ ] = eˣ
d/dx[ ln x ] = 1/x (x > 0)
d/dx[ aˣ ] = aˣ · ln a (a > 0, a ≠ 1)
d/dx[ logₐ x ] = 1/(x · ln a) (x > 0)
Key points:
eˣ is its own derivative. The base e ≈ 2.71828 is defined precisely so that y = eˣ has slope equal to its own height everywhere. No other base does this.aˣ needs an extra ln a. Since aˣ = e^(x ln a), the chain rule gives aˣ · ln a. When a = e, ln e = 1 and the factor disappears. Forgetting ln a is the signature aˣ error.1/ln a factor because logₐ x = (ln x)/(ln a).On the exam these functions almost never appear alone — they wrap an inner function g(x):
d/dx[ e^(g(x)) ] = e^(g(x)) · g'(x)
d/dx[ ln(g(x)) ] = g'(x) / g(x)
d/dx[ a^(g(x)) ] = a^(g(x)) · ln a · g'(x)
The ln(g) rule deserves a second look: outer derivative 1/g times inner derivative g' gives g'/g. The g' on top is the part students forget — writing d/dx[ln(x² + 1)] = 1/(x² + 1) (missing the 2x) is a classic dropped-chain-rule mistake. Quick examples:
d/dx[ e^(3x) ] = 3e^(3x)
d/dx[ e^(x²) ] = 2x·e^(x²)
d/dx[ ln(x² + 1) ]= 2x/(x² + 1)
d/dx[ ln(cos x) ] = −sin x / cos x = −tan x
Suppose f is differentiable and one-to-one. Because undoing then redoing returns the start, f(f⁻¹(x)) = x; differentiating both sides with the chain rule and solving gives:
(f⁻¹)'(b) = 1 / f'(f⁻¹(b)) provided f'(f⁻¹(b)) ≠ 0
In words: the derivative of the inverse at b is the reciprocal of f' at the matching input a = f⁻¹(b). The graph of f⁻¹ is the mirror image of f across y = x, and reflection turns a slope into its reciprocal.
Three-step recipe for (f⁻¹)'(b):
1. Find a = f⁻¹(b) — the input where f(a) = b.
2. Compute f'(a).
3. Take the reciprocal: (f⁻¹)'(b) = 1/f'(a).
The only trap is bookkeeping: you must evaluate f' at a, not at b. For f(x) = x³ + x + 1, since f(1) = 3, we have f⁻¹(3) = 1; then f'(x) = 3x² + 1 gives f'(1) = 4, so (f⁻¹)'(3) = 1/4 — with no need to invert the cubic.
d/dx[ arcsin x ] = 1/√(1 − x²) d/dx[ arcsin u ] = u'/√(1 − u²)
d/dx[ arccos x ] = −1/√(1 − x²) d/dx[ arccos u ] = −u'/√(1 − u²)
d/dx[ arctan x ] = 1/(1 + x²) d/dx[ arctan u ] = u'/(1 + u²)
Two memory hooks: the "co-" function is the negative (arccos is minus arcsin), and there are only two shapes here — 1/√(1 − x²) and 1/(1 + x²). Each chain-rule version multiplies by the inside derivative u'.
Differentiate f once for f', again for the second derivative f'', again for f''', and so on. After three primes the notation switches to a parenthesized superscript: f⁽⁴⁾(x), f⁽⁵⁾(x), …, f⁽ⁿ⁾(x). The parentheses matter: f⁴(x) would mean [f(x)]⁴, while f⁽⁴⁾(x) means the fourth derivative. Equivalent notations for the second derivative: f''(x), y'', d²y/dx².
For a degree-n polynomial, each derivative lowers the degree by one, so the nth derivative is a constant and every derivative after that is 0. If f(x) = x⁵ − 4x³ + 2x² − 7x + 9, then f⁽⁵⁾(x) = 5! = 120 and f⁽⁶⁾(x) = 0.
The sin/cos cycle repeats with period 4:
sin x → cos x → −sin x → −cos x → sin x → …
To find a high-order derivative of sin x, divide the order by 4 and use the remainder (0 → sin x, 1 → cos x, 2 → −sin x, 3 → −cos x). The 59th derivative: 59 = 4·14 + 3, remainder 3, so f⁽⁵⁹⁾(x) = −cos x.
The fixed point eˣ: every derivative of eˣ is eˣ. With a constant inside, factors compound: if f(x) = e^(kx), then f⁽ⁿ⁾(x) = kⁿ e^(kx). So d³/dx³[e^(3x)] = 3³ e^(3x) = 27e^(3x). Logarithms fall into a pattern too: from f(x) = ln x, f'(x) = 1/x, f''(x) = −1/x², f'''(x) = 2/x³.
If s(t) is position, then velocity is v(t) = s'(t) and acceleration is a(t) = s''(t) — the second derivative of position. A frequent slip is reporting velocity as acceleration; acceleration requires differentiating twice.
eˣ is its own derivative; aˣ carries an extra ln a; ln x gives 1/x and logₐ x gives 1/(x ln a).d/dx[e^g] = e^g·g' and d/dx[ln g] = g'/g — the g' on top is the most-forgotten factor.(f⁻¹)'(b) = 1/f'(f⁻¹(b)) — reciprocal of f' at the matching input, never at b.arcsin' = 1/√(1−x²), arctan' = 1/(1+x²); the "co-" versions are negative.f⁽ⁿ⁾ means differentiate n times; a degree-n polynomial dies after n derivatives.sin/cos with period 4, and every derivative of e^(kx) multiplies by another k.d/dx[e^(4x)] =1. C) 4e^(4x). Chain rule: e^(4x)·4. Distractors: A) forgot the inside derivative 4. B) wrongly applied the power rule to an exponential. D) divided by 4 instead of multiplying. E) squared the inside factor (used 4·4). Fix: d/dx[e^(ax)] = a·e^(ax); the base e self-differentiates, and the inside factor a multiplies.
d/dx[ln(x² + 4)] =2. E) 2x/(x² + 4). The ln(g) rule with g = x² + 4, g' = 2x: g'/g. Distractors: A) dropped the inside derivative 2x. B) kept only g' and lost the 1/g. C) inverted the inside. D) forgot the x in the numerator (used g' = 2 by mistake). Fix: d/dx[ln g] = g'/g — always put the inside derivative on top.
d/dx[3ˣ] =3. D) 3ˣ·ln 3. The aˣ rule: d/dx[aˣ] = aˣ ln a. Distractors: A) forgot ln 3 — the signature aˣ error, treating base 3 like base e. B) applied the power rule to a constant base. C) divided by ln 3 instead of multiplying. E) invented a factor of x. Fix: only base e self-differentiates; every other base multiplies by ln a.
f satisfies f(1) = 4 and f'(1) = 7. If g = f⁻¹, then g'(4) =4. B) 1/7. Inverse formula: since f(1) = 4, f⁻¹(4) = 1, so g'(4) = 1/f'(1) = 1/7. Distractors: A) 7 forgot to take the reciprocal. C) 1/4 took the reciprocal of the output 4 instead of f'. D) 4 reported the value b. E) 7/4 mixed the two numbers. Fix: (f⁻¹)'(b) = 1/f'(a) where f(a) = b; find the matching input first, then reciprocate f' there.
f(x) = e^(3x), then f'''(x) =5. C) 27e^(3x). Each derivative pulls down a factor of 3: f''' = 3³ e^(3x) = 27e^(3x). Distractors: A) ignored the chain rule entirely. B) applied it once. D) applied it twice (9). E) applied it four times (81). Fix: for e^(kx), the nth derivative is kⁿ e^(kx) — count the derivatives to count the factors of k.
sin x is6. A) −cos x. The cycle sin → cos → −sin → −cos has period 4; 59 = 4·14 + 3, remainder 3 → −cos x. Distractors: B) remainder 1. C) remainder 0. D) remainder 2. E) the derivative exists — sine is differentiable forever. Fix: reduce the order mod 4 and read the cycle position.
d/dx[arctan x] =7. A) 1/(1 + x²). Standard inverse-trig derivative. Distractors: B) is the arcsin derivative. C) is arccot (wrong sign). D) is arccos. E) confuses 1 + x² with 1 − x². Fix: memorize arctan' = 1/(1 + x²) and arcsin' = 1/√(1 − x²) — the 1 + x² shape has no square root.
d/dx[arcsin(3x)] =8. A) 3/√(1 − 9x²). With u = 3x, u' = 3: u'/√(1 − u²) = 3/√(1 − 9x²). Distractors: B) forgot the chain-rule factor 3. C) mishandled u² = 9x² (wrote 3x²). D) used the arccos sign. E) used the arctan shape. Fix: for arcsin(u), multiply by u' and square the whole inside: u² = (3x)² = 9x².
The increasing differentiable function f has the values below. If g = f⁻¹, then g'(8) =
| x | f(x) | f'(x) |
|---|---|---|
| 2 | 8 | 5 |
| 3 | 11 | 2 |
| 4 | 15 | 9 |
9. E) 1/5. Find a = f⁻¹(8): since f(2) = 8, a = 2, so g'(8) = 1/f'(2) = 1/5. Distractors: A) 5 forgot the reciprocal. B) 1/2 used 1/a instead of 1/f'(a). C) 1/8 took the reciprocal of the output 8. D) 2 reported the input a. Fix: the inverse-derivative answer is 1/f'(a), where a is the row whose f-value equals the target.
s(t) = t³ − 6t² + 9t + 1. Its acceleration at t = 4 is10. D) 12. v(t) = 3t² − 12t + 9, a(t) = 6t − 12; at t = 4, a(4) = 24 − 12 = 12. Distractors: A) 0 used a(2). B) 6 computed 6t only at a wrong point. C) 9 reported a constant from v. E) −6 used a(1). Fix: acceleration is the second derivative of position; differentiate twice, then substitute.
d/dx[x·eˣ] =11. E) (x + 1)eˣ. Product rule: (1)eˣ + x·eˣ = (x + 1)eˣ. Distractors: A) kept only the first term. B) kept only the second. C) sign error (should be +1). D) wrongly applied the power rule to eˣ. Fix: d/dx[x·eˣ] needs the product rule, and eˣ differentiates to itself.
f(x) = ln x, then f'''(x) =12. B) 2/x³. f'(x) = x⁻¹, f''(x) = −x⁻² = −1/x², f'''(x) = 2x⁻³ = 2/x³. Distractors: A) dropped the factor 2. C) is f'', stopping one derivative early. D) has the wrong sign. E) is f⁽⁴⁾, one derivative too far. Fix: differentiate the power form x⁻¹ step by step, tracking both the falling exponent and the accumulating coefficient.
1. C) 4e^(4x). Chain rule: e^(4x)·4. Distractors: A) forgot the inside derivative 4. B) wrongly applied the power rule to an exponential. D) divided by 4 instead of multiplying. E) squared the inside factor (used 4·4). Fix: d/dx[e^(ax)] = a·e^(ax); the base e self-differentiates, and the inside factor a multiplies.
2. E) 2x/(x² + 4). The ln(g) rule with g = x² + 4, g' = 2x: g'/g. Distractors: A) dropped the inside derivative 2x. B) kept only g' and lost the 1/g. C) inverted the inside. D) forgot the x in the numerator (used g' = 2 by mistake). Fix: d/dx[ln g] = g'/g — always put the inside derivative on top.
3. D) 3ˣ·ln 3. The aˣ rule: d/dx[aˣ] = aˣ ln a. Distractors: A) forgot ln 3 — the signature aˣ error, treating base 3 like base e. B) applied the power rule to a constant base. C) divided by ln 3 instead of multiplying. E) invented a factor of x. Fix: only base e self-differentiates; every other base multiplies by ln a.
4. B) 1/7. Inverse formula: since f(1) = 4, f⁻¹(4) = 1, so g'(4) = 1/f'(1) = 1/7. Distractors: A) 7 forgot to take the reciprocal. C) 1/4 took the reciprocal of the output 4 instead of f'. D) 4 reported the value b. E) 7/4 mixed the two numbers. Fix: (f⁻¹)'(b) = 1/f'(a) where f(a) = b; find the matching input first, then reciprocate f' there.
5. C) 27e^(3x). Each derivative pulls down a factor of 3: f''' = 3³ e^(3x) = 27e^(3x). Distractors: A) ignored the chain rule entirely. B) applied it once. D) applied it twice (9). E) applied it four times (81). Fix: for e^(kx), the nth derivative is kⁿ e^(kx) — count the derivatives to count the factors of k.
6. A) −cos x. The cycle sin → cos → −sin → −cos has period 4; 59 = 4·14 + 3, remainder 3 → −cos x. Distractors: B) remainder 1. C) remainder 0. D) remainder 2. E) the derivative exists — sine is differentiable forever. Fix: reduce the order mod 4 and read the cycle position.
7. A) 1/(1 + x²). Standard inverse-trig derivative. Distractors: B) is the arcsin derivative. C) is arccot (wrong sign). D) is arccos. E) confuses 1 + x² with 1 − x². Fix: memorize arctan' = 1/(1 + x²) and arcsin' = 1/√(1 − x²) — the 1 + x² shape has no square root.
8. A) 3/√(1 − 9x²). With u = 3x, u' = 3: u'/√(1 − u²) = 3/√(1 − 9x²). Distractors: B) forgot the chain-rule factor 3. C) mishandled u² = 9x² (wrote 3x²). D) used the arccos sign. E) used the arctan shape. Fix: for arcsin(u), multiply by u' and square the whole inside: u² = (3x)² = 9x².
9. E) 1/5. Find a = f⁻¹(8): since f(2) = 8, a = 2, so g'(8) = 1/f'(2) = 1/5. Distractors: A) 5 forgot the reciprocal. B) 1/2 used 1/a instead of 1/f'(a). C) 1/8 took the reciprocal of the output 8. D) 2 reported the input a. Fix: the inverse-derivative answer is 1/f'(a), where a is the row whose f-value equals the target.
10. D) 12. v(t) = 3t² − 12t + 9, a(t) = 6t − 12; at t = 4, a(4) = 24 − 12 = 12. Distractors: A) 0 used a(2). B) 6 computed 6t only at a wrong point. C) 9 reported a constant from v. E) −6 used a(1). Fix: acceleration is the second derivative of position; differentiate twice, then substitute.
11. E) (x + 1)eˣ. Product rule: (1)eˣ + x·eˣ = (x + 1)eˣ. Distractors: A) kept only the first term. B) kept only the second. C) sign error (should be +1). D) wrongly applied the power rule to eˣ. Fix: d/dx[x·eˣ] needs the product rule, and eˣ differentiates to itself.
12. B) 2/x³. f'(x) = x⁻¹, f''(x) = −x⁻² = −1/x², f'''(x) = 2x⁻³ = 2/x³. Distractors: A) dropped the factor 2. C) is f'', stopping one derivative early. D) has the wrong sign. E) is f⁽⁴⁾, one derivative too far. Fix: differentiate the power form x⁻¹ step by step, tracking both the falling exponent and the accumulating coefficient.