Difficulty: Intermediate | Prerequisites: Power rule, properties of ln and e, compound interest formulas
This section asks you to differentiate functions involving ln(x) and e^x. The key skill is recognising when to simplify the expression with log rules before differentiating, and when to apply the derivative rules directly. You will also use these derivatives to find tangent lines and to interpret rates of change in applied problems (investment growth, drug clearance). This connects the algebra of logarithms from the previous section to the calculus that drives the rest of the exam.
The derivative of ln(x) is 1/x. The derivative of e^(kx) is k * e^(kx). Before differentiating, simplify using log properties (expand products, quotients and powers). For tangent lines, find the slope via f'(a) and the point via f(a), then use y = mx + b.
d/dx [ln(x)] = 1/x
The derivative of the natural logarithm. Only valid for x > 0.
d/dx [e^x] = e^x
The exponential function is its own derivative. This is one of the defining properties of e.
d/dx [e^(kx)] = k * e^(kx)
When the exponent is a constant multiple of x, the constant "comes down" as a multiplier. This is a direct application of the chain rule.
Logarithm properties (for simplification before differentiating)
ln(ab) = ln a + ln b
ln(a/b) = ln a - ln b
ln(a^n) = n * ln a
These let you break a complicated log expression into simpler terms you can differentiate one at a time.
Tangent line
The line that touches a curve at exactly one point and has the same slope as the curve at that point. Found using: slope = f'(a), point = (a, f(a)), then y - f(a) = f'(a)(x - a).
In simple terms, the line that "just grazes" the curve at a given x-value.
Instantaneous rate of change
The derivative evaluated at a specific point. In applied problems, it tells you how quickly a quantity (investment balance, drug concentration) is changing at that exact moment.
The exam instructions say: "clearly separate the algebra from the derivative." This means you should rewrite the function using log properties or exponent rules before you differentiate.
Example (Problem 18): y = 6 ln(x) - x³ + 2∛x. Rewrite the cube root: y = 6 ln(x) - x³ + 2x^(1/3). Now differentiate term by term: y' = 6/x - 3x² + (2/3)x^(-2/3).
Example (Problem 19): y = 4√x + ln(x⁹). Rewrite: y = 4x^(1/2) + 9 ln(x). Differentiate: y' = 2/√x + 9/x.
Example (Problem 20): y = ln(4/x³). Expand: y = ln 4 - 3 ln x. Differentiate: y' = 0 - 3/x = -3/x. (ln 4 is a constant, its derivative is zero.)
Example (Problem 21): y = 3x^e - 2e^(5x) - ln(3x). Rewrite the last term: ln(3x) = ln 3 + ln x. So y = 3x^e - 2e^(5x) - ln 3 - ln x. Differentiate: y' = 3e * x^(e-1) - 10e^(5x) - 1/x. (ln 3 is a constant.)
d/dx [ln(x)] = 1/x
d/dx [ln(kx)] = 1/x (because ln(kx) = ln k + ln x, and ln k is a constant)
d/dx [e^x] = e^x
d/dx [e^(kx)] = k * e^(kx)
d/dx [x^n] = n * x^(n-1) (power rule, works for any constant n, including fractions like 1/3 or irrational numbers like e)
The method is always the same three steps:
Find the y-coordinate: y₀ = f(a).
Find the slope: m = f'(a).
Write the equation: y - y₀ = m(x - a), then rearrange to slope-intercept form y = mx + b.
Example (Problem 22): f(x) = e^x + 5x - 4 at x = 0. f(0) = 1 + 0 - 4 = -3. f'(x) = e^x + 5, so f'(0) = 1 + 5 = 6. Tangent line: y - (-3) = 6(x - 0), so y = 6x - 3.
Example (Problem 23): f(x) = 1 + ln(x⁴) = 1 + 4 ln x at x = e. f(e) = 1 + 4(1) = 5. f'(x) = 4/x, so f'(e) = 4/e. Tangent line: y - 5 = (4/e)(x - e), so y = (4/e)x + 5 - 4 = (4/e)x + 1.
Investment growth (Problem 24): A = 25000e^(0.084t). The instantaneous rate of change is A'(t) = 25000(0.084)e^(0.084t) = 2100e^(0.084t).
At t = 2: A'(2) = 2100e^(0.168) = $2,484.17. This means after 2 years, the account balance is increasing at a rate of about $2,484 per year.
When balance = $30,000: set 30000 = 25000e^(0.084t), so e^(0.084t) = 1.2. Then A'(t) = 2100 * 1.2 = $2,520.
Drug concentration (Problem 25): C(t) = 5e^(-0.3t). C'(t) = 5(-0.3)e^(-0.3t) = -1.5e^(-0.3t).
At t = 1: C'(1) = -1.5e^(-0.3) = -1.11 mg/mL per hour. The concentration is decreasing.
At t = 5: C'(5) = -1.5e^(-1.5) = -0.33 mg/mL per hour. Still decreasing, but more slowly. Over time, the rate of decrease slows because there is less drug left to clear.
\frac{d}{dx}[\ln(x)] = \frac{1}{x}\frac{d}{dx}[e^{kx}] = k \cdot e^{kx}\frac{d}{dx}[x^n] = n \cdot x^{n-1} \quad \text{(for any constant } n \text{)}Tangent line at x = a:
y = f'(a)(x - a) + f(a)Log expansion rules:
\ln(ab) = \ln a + \ln b \qquad \ln\!\left(\frac{a}{b}\right) = \ln a - \ln b \qquad \ln(a^n) = n \ln aForgetting that ln(constant) is just a number. ln 3, ln 4, ln 2 are all constants. Their derivatives are zero. Students sometimes try to differentiate them.
Differentiating x^e as if e were a variable. In the expression x^e, the exponent e is the constant 2.718... Use the power rule: d/dx[x^e] = e * x^(e-1). Do not use the exponential rule here.
Confusing d/dx[e^x] with d/dx[x^e]. The first is e^x (exponential rule). The second is e * x^(e-1) (power rule). They look similar but are entirely different operations.
Not simplifying before differentiating. If you differentiate ln(4/x³) directly without expanding, you will need the chain rule and quotient rule. Expanding to ln 4 - 3 ln x first makes it a one-step problem.
Slope-intercept errors in tangent lines. Students find the slope correctly but then write y = mx without the intercept, or confuse the point (a, f(a)) with (0, f(0)). Always plug the point back in to solve for b.
⚠️ The exam will ask you to show the algebra step separately from the derivative. Write f(x) = [simplified form], then on a new line write f'(x) = [derivative]. Credit depends on this.
⚠️ Tangent line problems will appear. Be ready to find slope, point and equation in slope-intercept form.
⚠️ Applied problems (investment rate of change, drug concentration rate of change) require interpreting the answer in context. State the units and what the sign means.
⚠️ Expanding ln expressions before differentiating is tested directly in problems 17 and 20.
Fill in the blank: d/dx [ln(5x)] = ___. (1/x, because ln(5x) = ln 5 + ln x, and d/dx[ln x] = 1/x.)
True or False: d/dx [e^(3x)] = e^(3x). (False. It is 3e^(3x).)
Fill in the blank: d/dx [x^e] = ___. (e * x^(e-1). Power rule, not exponential rule.)
True or False: To find a tangent line at x = a, you need both f(a) and f'(a). (True.)
Fill in the blank: If C(t) = 5e^(-0.3t), then C'(t) = ___. (-1.5e^(-0.3t))
Q: Expand ln(xy³/z) using log properties.
A: ln x + 3 ln y - ln z.
Q: Differentiate y = 6 ln(x) - x³ + 2x^(1/3).
A: y' = 6/x - 3x² + (2/3)x^(-2/3).
Q: Differentiate y = ln(4/x³).
A: Algebra first: y = ln 4 - 3 ln x. Then y' = -3/x.
Q: Differentiate y = 3x^e - 2e^(5x) - ln(3x).
A: Rewrite: y = 3x^e - 2e^(5x) - ln 3 - ln x. Then y' = 3e * x^(e-1) - 10e^(5x) - 1/x.
Q: Find the tangent line to f(x) = e^x + 5x - 4 at x = 0.
A: f(0) = 1 + 0 - 4 = -3. f'(x) = e^x + 5, f'(0) = 6. Tangent: y = 6x - 3.
Q: C(t) = 5e^(-0.3t). Find C'(1) and C'(5). Interpret.
A: C'(t) = -1.5e^(-0.3t). C'(1) = -1.11 mg/mL per hour (concentration falling). C'(5) = -0.33 mg/mL per hour (still falling, but more slowly as less drug remains).
Q: A = 25000e^(0.084t). Find the rate of change when the balance reaches $30,000.
A: A'(t) = 2100e^(0.084t). When A = 30000: 30000/25000 = e^(0.084t) = 1.2. So A'(t) = 2100(1.2) = $2,520 per year.
The derivative rules here are the building blocks for the product rule and quotient rule problems that follow (problems 26-37 on the review). If you can differentiate ln(x) and e^x confidently on their own, combining them with other functions via product and quotient rules becomes mechanical.
The tangent-line technique reappears in the next set of problems (34, where you find the tangent to a quotient) and in the horizontal-tangent problem (36, where you set the derivative equal to zero).
derivative of ln x, derivative of e^x, logarithmic differentiation, exponential differentiation, tangent line, slope-intercept form, instantaneous rate of change, drug concentration, continuous growth, MAC 2233, business calculus, Exam 2, FIU, log properties, algebra before calculus, power rule with fractional exponents