Lesson 14: Derivatives of Inverse & Exponential/Log Functions

(a) Opening Question

The function f(x) = x³ + x + 1 is always increasing (its derivative 3x² + 1 is never zero), so it has an inverse f⁻¹. Notice that f(1) = 1 + 1 + 1 = 3, which means f⁻¹(3) = 1.

Now the real question: what is the slope of f⁻¹ at the point where x = 3?

You could try to solve y = x³ + x + 1 for x in terms of y — but that's a cubic, and there's no clean formula. So a direct approach is hopeless.

Here's the key idea. The graph of f⁻¹ is the mirror image of the graph of f across the line y = x. Reflecting a line across y = x flips its slope into the reciprocal. The graph of f passes through (1, 3) with slope f'(1) = 3(1)² + 1 = 4. So the graph of f⁻¹ passes through the mirror point (3, 1) with slope 1/4.

That single reciprocal relationship — slope of the inverse = reciprocal of the original slope — is the engine of the whole first half of this lesson.

(b) Core Concepts

The inverse-function derivative formula

Suppose f is differentiable and one-to-one, with inverse f⁻¹. By definition, undoing f then f returns you to the start:

f(f⁻¹(x)) = x

Differentiate both sides with respect to x. The right side gives 1. The left side is a composite, so the Chain Rule applies:

f'(f⁻¹(x)) · (f⁻¹)'(x) = 1

Solve for the thing we want, (f⁻¹)'(x):

Inverse-Function Derivative. If f is differentiable and one-to-one, and f⁻¹(b) = a (equivalently f(a) = b), then

`

(f⁻¹)'(b) = ────────────── = ────────

f'(f⁻¹(b)) f'(a)

`

provided f'(a) ≠ 0.

In words: the derivative of the inverse at b is the reciprocal of the derivative of f at the matching input a. The only trick is bookkeeping — you must evaluate f' at a = f⁻¹(b), not at b.

The three-step recipe. To find (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).

Mini-example. For f(x) = x³ + x + 1 from the Opening Question, find (f⁻¹)'(3).

• Step 1: a = f⁻¹(3). Since f(1) = 3, we have a = 1.
• Step 2: f'(x) = 3x² + 1, so f'(1) = 4.
• Step 3: (f⁻¹)'(3) = 1/f'(1) = 1/4. ✓ (Matches the reflection argument.)

The exponential and logarithmic derivative table

These four results are worth memorizing cold. The first is the most remarkable fact in calculus.

Exponential and Logarithmic Derivatives.

`

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)

`

A few things to internalize:

• eˣ is its own derivative. The base e ≈ 2.71828 is defined precisely so that the curve y = eˣ has slope equal to its own height at every point. No other base does this.
• aˣ needs an extra ln a. Since aˣ = e^(x ln a), the Chain Rule gives d/dx[e^(x ln a)] = e^(x ln a) · ln a = aˣ · ln a. When a = e, ln e = 1 and the factor disappears — that's why eˣ is special. Forgetting ln a is the #1 error with aˣ.
• ln x and logₐ x. The natural log differentiates to 1/x. A general log carries a 1/ln a factor because logₐ x = (ln x)/(ln a), and 1/(ln a) is just a constant multiplier.

eˣ and ln x are inverses

The relationship e^(ln x) = x and ln(eˣ) = x says these two functions undo each other. We can even recover d/dx[ln x] = 1/x from the inverse formula. Let f(x) = eˣ, so f⁻¹(x) = ln x. Then

(ln x)' = (f⁻¹)'(x) = 1/f'(f⁻¹(x)) = 1/e^(ln x) = 1/x

because f'(u) = eᵘ and f⁻¹(x) = ln x, so f'(f⁻¹(x)) = e^(ln x) = x. The inverse formula and the derivative table are the same idea wearing two outfits.

Composites: the Chain Rule with exp and log

On the AP exam, eˣ and ln x almost never appear alone — they wrap around an inner function g(x). The Chain Rule ("derivative of outside, keeping the inside, times derivative of the inside") gives:

Chain Rule with exp/log.

`

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(x)) rule is worth a special look: the outside derivative 1/(inside) times the inside derivative g'(x) gives g'(x)/g(x). The g'(x) 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.

Mini-examples.

d/dx[ e^(3x) ]    = e^(3x) · 3        = 3e^(3x)
d/dx[ e^(x²) ]    = e^(x²) · 2x       = 2x·e^(x²)
d/dx[ ln(x²+1) ]  = (2x)/(x²+1)
d/dx[ ln(cos x) ] = (−sin x)/(cos x)  = −tan x

Logarithmic differentiation (the idea, briefly)

When a function is a messy product, quotient, or — especially — a variable base raised to a variable power like xˣ, taking the natural log first turns multiplication into addition and exponents into multipliers. The headline case:

y = xˣ   ⟹   ln y = x·ln x   ⟹   (1/y)·y' = ln x + 1   ⟹   y' = xˣ(ln x + 1)

You differentiate ln y on the left (giving y'/y by the Chain Rule) and the simplified right side, then multiply back by y. You won't need heavy log-differentiation on the AB exam, but recognizing that a power tower like xˣ is neither the power rule nor the aˣ rule — both require a constant somewhere — is exactly the kind of distinction AP likes to test.

Calculator note. This lesson is Mixed. Hand-differentiation is expected on non-calculator items; on [CALC] items you may confirm a numerical derivative with nDeriv. Example: TI-84: MATH → 8:nDeriv( → nDeriv(e^(2X), X, 0) returns 2, confirming d/dx[e^(2x)] = 2e^(2x) equals 2 at x = 0. Use nDeriv to check, never to replace the rule.

(c) Worked Examples

Example 1 — Inverse derivative from a table (foundation) · [NO CALC]

Problem. The differentiable, increasing function f has the values below. Let g = f⁻¹. Find g'(5).

Strategy. Use (f⁻¹)'(b) = 1/f'(f⁻¹(b)) with b = 5. First find the input a where f(a) = 5.

Solution.

g'(5) = (f⁻¹)'(5) = 1 / f'(f⁻¹(5))

From the table, f(2) = 5, so f⁻¹(5) = 2. Then f'(2) = 4:

g'(5) = 1 / f'(2) = 1/4

Answer. g'(5) = 1/4.

Justification. The matching input is a = f⁻¹(5) = 2 because f(2) = 5; the inverse-derivative formula then requires f' evaluated at that input, f'(2) = 4, giving the reciprocal 1/4. Reading f'(5) off the table would be the wrong evaluation point (and 5 isn't even an input row).

Example 2 — e^(g(x)) and ln(g(x)) chain rule (procedural) · [NO CALC]

Problem. Differentiate (i) y = e^(x² − 3x) and (ii) y = ln(3x² − 2).

Strategy. Both are composites. Apply the Chain Rule: outside derivative times inside derivative.

Solution.

(i) Outside eᵘ differentiates to eᵘ; inside u = x² − 3x has u' = 2x − 3.

y' = e^(x² − 3x) · (2x − 3) = (2x − 3)·e^(x² − 3x)

(ii) Outside ln u differentiates to 1/u; inside u = 3x² − 2 has u' = 6x.

y' = (6x)/(3x² − 2)

Answer. (i) (2x − 3)e^(x² − 3x); (ii) 6x/(3x² − 2).

Justification. In (ii) the numerator 6x is the inside derivative g'(x); the ln(g) rule is g'/g, so dropping the 6x and writing 1/(3x² − 2) would be the missing-chain-rule error.

Example 3 — aˣ and logₐ x (procedural) · [NO CALC]

Problem. Differentiate (i) y = 5^x, (ii) y = log₂ x, and (iii) y = 3^(x²+1).

Strategy. Use d/dx[aˣ] = aˣ ln a and d/dx[logₐ x] = 1/(x ln a). Part (iii) layers the Chain Rule on top of the aˣ rule.

Solution.

(i)   y' = 5^x · ln 5

(ii)  y' = 1 / (x · ln 2)

(iii) y' = 3^(x²+1) · ln 3 · (2x) = 2x·ln 3 · 3^(x²+1)

Answer. (i) 5^x ln 5; (ii) 1/(x ln 2); (iii) 2x·ln 3·3^(x²+1).

Justification. Each non-e base carries a ln a factor (ln 5, ln 2, ln 3). In (iii) the Chain Rule supplies the additional g'(x) = 2x. Omitting ln 3 — treating 3^(x²+1) as if it were the power rule on a variable — is the signature aˣ mistake.

Example 4 — Product with eˣ, evaluated and interpreted (AP level) · [CALC to confirm]

Problem. Let h(x) = x²·e^(3x). Find h'(x), then find the slope of the tangent line to h at x = 0.

Strategy. A genuine product → Product Rule, where the second factor e^(3x) itself needs the Chain Rule. Then evaluate h'(0).

Solution. With f = x² (f' = 2x) and p = e^(3x) (p' = 3e^(3x) by the Chain Rule):

h'(x) = f'·p + f·p'
      = (2x)·e^(3x) + x²·(3e^(3x))
      = e^(3x)·(2x + 3x²)
      = x(3x + 2)·e^(3x)

At x = 0:

h'(0) = 0·(0 + 2)·e^0 = 0

Answer. h'(x) = x(3x + 2)e^(3x); the tangent slope at x = 0 is 0.

Justification. The Product Rule keeps one factor undifferentiated per term; the e^(3x) factor's derivative carries its own Chain-Rule factor of 3. Factoring out the common e^(3x) (never zero) leaves 2x + 3x², which is 0 at x = 0. Check with nDeriv: nDeriv(X²·e^(3X), X, 0) returns 0. ✓

(d) Common Mistakes

Writing d/dx[eˣ] = x·e^(x−1). Students apply the power rule to eˣ. ✗ The power rule is for a variable base, constant exponent (xⁿ). Here the base is constant (e) and the exponent is the variable — a totally different rule. The correct (and remarkable) answer is d/dx[eˣ] = eˣ.

Forgetting ln a for aˣ. Writing d/dx[2ˣ] = 2ˣ by analogy with eˣ. ✗ Only base e self-differentiates. For any other base, d/dx[aˣ] = aˣ·ln a. Correct: d/dx[2ˣ] = 2ˣ ln 2.

Dropping g'(x) in d/dx[ln(g(x))]. Writing d/dx[ln(x²+1)] = 1/(x²+1). ✗ The Chain Rule requires the inside derivative on top: d/dx[ln(g)] = g'/g, so the answer is 2x/(x²+1). The g' is the most-forgotten piece in the whole lesson.

Evaluating f' at the wrong point for the inverse formula. Computing (f⁻¹)'(b) as 1/f'(b). ✗ The formula is 1/f'(f⁻¹(b)) — you must first find a = f⁻¹(b) and evaluate f' there. Plug into f'(a), never f'(b).

Trying to invert f algebraically instead of using the reciprocal formula. Spending the whole problem solving y = f(x) for x. ✗ The inverse-derivative formula (f⁻¹)'(b) = 1/f'(a) is designed so you never need a formula for f⁻¹ — just the matching point and f' there. For a cubic like x³ + x + 1, inverting is impossible anyway.

(e) Practice Problems

Attempt before checking section (g). [NO CALC] unless marked [CALC].

Differentiate y = ln(3x² − 2). Show your setup.

Differentiate y = x²·ln x. (Hint: Product Rule.)

For S(t) = 50·e^(0.2t) (a subscriber count, in thousands, t years after launch), find S'(5) and write one sentence interpreting it with correct units.

(justify) A student computes d/dx[ln(cos x)] = 1/(cos x). Compute the correct derivative and, in one sentence, identify the rule the student dropped.

(justify) The function f(x) = x⁵ + 2x + 1 is one-to-one with f(1) = 4. Without finding a formula for f⁻¹, compute (f⁻¹)'(4) and justify, in one sentence, why you do not need to invert f.

(f) AP Exam Focus — Free Response

> Free Response Question — 9 points total

> Parts (a), (c), (d): No calculator. Part (b): Calculator permitted.

>

> A drug is administered to a patient. For 0 ≤ t ≤ 8, the concentration of the drug in the patient's bloodstream, in milligrams per liter (mg/L), is modeled by

> `

> C(t) = 20·e^(−0.5t)

> `

> where t is measured in hours after the dose.

>

> Separately, a nurse records the differentiable, increasing function D, the cumulative dose delivered (in mg) as a function of pump setting s. Selected values appear in the table.

>

> | s | D(s) | D'(s) |

> |---|---|---|

> | 1 | 2 | 3 |

> | 2 | 5 | 4 |

> | 3 | 10 | 6 |

>

> (a) (3 points) Find C'(2), including units, and interpret its meaning in the context of the problem.

>

> (b) (2 points) Find the time t, for 0 ≤ t ≤ 8, at which the concentration reaches 5 mg/L. Show the equation that leads to your answer.

>

> (c) (2 points) Write an equation for the line tangent to the graph of C at t = 2, and use it to approximate C(2.1).

>

> (d) (2 points) Let D⁻¹ be the inverse of the function D. Find (D⁻¹)'(5). Show the setup that leads to your answer.

MODEL SOLUTION

(a) C'(2) with units and interpretation — 3 points

Differentiate using the Chain Rule (d/dx[e^(g)] = e^(g)·g'):

C'(t) = 20·e^(−0.5t)·(−0.5) = −10·e^(−0.5t)

Evaluate at t = 2:

C'(2) = −10·e^(−0.5·2) = −10·e^(−1) = −10/e ≈ −3.679 mg/L per hour

Interpretation: At t = 2 hours, the drug concentration in the bloodstream is decreasing at a rate of about 3.679 mg/L per hour. (The negative sign indicates the concentration is falling.)

(b) Time when C(t) = 5 — 2 points

Set the model equal to 5 and solve:

20·e^(−0.5t) = 5
e^(−0.5t) = 1/4
−0.5t = ln(1/4)
t = −2·ln(1/4) = 2·ln 4 = ln 16 ≈ 2.773 hours

t ≈ 2.773 hours.

(c) Tangent line at t = 2; approximate C(2.1) — 2 points

The point: C(2) = 20·e^(−1) = 20/e ≈ 7.358. The slope: C'(2) = −10/e ≈ −3.679 (from part (a)).

Tangent line through (2, 20/e) with slope −10/e:

y = 20/e − (10/e)(t − 2)        [≈ 7.358 − 3.679(t − 2)]

Approximate C(2.1):

C(2.1) ≈ 7.358 − 3.679(2.1 − 2) = 7.358 − 3.679(0.1) ≈ 6.990

Tangent: y = 20/e − (10/e)(t − 2); C(2.1) ≈ 6.990 mg/L.

(d) (D⁻¹)'(5) — 2 points

By the inverse-function derivative formula, (D⁻¹)'(b) = 1/D'(D⁻¹(b)). With b = 5: from the table, D(2) = 5, so D⁻¹(5) = 2. Then D'(2) = 4:

(D⁻¹)'(5) = 1/D'(D⁻¹(5)) = 1/D'(2) = 1/4

(D⁻¹)'(5) = 1/4.

SCORING COMMENTARY — where students lose points

Part (a) — 3 pts: 1 pt for the correct derivative C'(t) = −10e^(−0.5t) (Chain Rule, including the −0.5 factor); 1 pt for the value ≈ −3.679 with units mg/L per hour; 1 pt for an interpretation that names the rate of decrease. Common loss: omitting the −0.5 chain factor (writing C'(t) = 20e^(−0.5t)) loses the derivative point; a bare number with no units loses the units point; "the concentration is −3.679" (confusing the rate with the value) loses the interpretation point.

Part (b) — 2 pts: 1 pt for the equation 20e^(−0.5t) = 5 (or e^(−0.5t) = 1/4); 1 pt for t = 2 ln 4 = ln 16 ≈ 2.773. Common loss: algebra slips when taking ln of both sides, or forgetting to divide by −0.5, which scrambles the answer.

Part (c) — 2 pts: 1 pt for a correct tangent equation using C(2) = 20/e and C'(2) = −10/e; 1 pt for the linear approximation C(2.1) ≈ 6.990. Common loss: finding the slope but never writing an actual line equation, or using C'(2) as the y-value instead of C(2). Decimal forms (7.358, −3.679) are accepted; show the line.

Part (d) — 2 pts: 1 pt for the correct setup (D⁻¹)'(5) = 1/D'(D⁻¹(5)) with D⁻¹(5) = 2 identified; 1 pt for the answer 1/4. Common loss: using 1/D'(5) — but 5 is not even an input row, and D'(5) is undefined here — earns 0; evaluating D' at the output instead of the matching input is the core error this part tests.

d/dx[ln(cos x)] = (−sin x)/(cos x) = −tan x