Differentiate exponential ($e^x$, $a^x$) and logarithmic ($\ln x$, $\log_b x$) functions, including composite functions via the chain rule

What this dot point is asking

VCAA wants you to differentiate exponential and logarithmic functions, including composite functions via the chain rule.

Standard derivatives

$e^x$ is the unique function (up to scaling) whose derivative equals itself.

Chain rule extensions

The absolute value extends the natural log to negative arguments (the derivative is the same form on both sides of zero).

Applications

For continuous decay $N = N_0 e^{-\lambda t}$:

The decay rate is proportional to the current amount; this is the differential-equation definition of exponential decay.

For population growth $P = P_0 e^{kt}$:

The growth rate is proportional to current population.

Combining rules: product, quotient and chain

Most exam derivatives mix an exponential or logarithm with another function, so you must choose the right combination of rules. A product such as $x^2 e^{2x}$ needs the product rule; a quotient such as $\dfrac{e^x}{x}$ needs the quotient rule; a composite such as $e^{3x^2 - 2}$ needs the chain rule. When two rules are needed (for example $x e^{\sin x}$), apply the product rule first, then the chain rule on the inner composite. A useful habit is to factor the common exponential out of the final answer, since $e^{u}$ is never zero and factoring makes solving $f'(x) = 0$ for stationary points immediate.

Logarithmic functions and their domain

The derivative $\dfrac{d}{dx}\ln x = \dfrac{1}{x}$ holds for $x > 0$, matching the domain of $\ln x$. Using $\ln|x|$ extends the rule to $x \ne 0$ with the same derivative $\dfrac{1}{x}$. For a composite $\ln(u(x))$, the chain rule gives $\dfrac{u'(x)}{u(x)}$, and the result is only valid where $u(x) > 0$. VCAA expects you to state the implied domain when differentiating logarithmic functions, because the gradient function inherits the restriction.

In one sentence

$\frac{d}{dx} e^x = e^x$, $\frac{d}{dx} \ln x = 1/x$, $\frac{d}{dx} a^x = a^x \ln a$, $\frac{d}{dx} \log_b x = 1/(x \ln b)$, with chain-rule extensions $\frac{d}{dx} e^u = u' e^u$ and $\frac{d}{dx} \ln u = u'/u$.

Examples in context

Example 1. Rate of cooling. A cup of coffee cools according to $T(t) = 20 + 60e^{-0.05t}$ degrees Celsius, $t$ minutes after pouring. The rate of cooling is $T'(t) = 60 \times (-0.05) e^{-0.05t} = -3e^{-0.05t}$ degrees per minute. At $t = 10$: $T'(10) = -3e^{-0.5} = -3 \times 0.6065 = -1.82$ degrees per minute (still cooling, but slowing).

Example 2. Marginal log-utility. An economic model uses $U(x) = 40\ln(x + 1)$ for the satisfaction from $x$ units of a good. The marginal utility is $U'(x) = \frac{40}{x + 1}$. At $x = 3$, $U'(3) = \frac{40}{4} = 10$ units of satisfaction per extra good; at $x = 9$, $U'(9) = \frac{40}{10} = 4$, illustrating diminishing returns as the chain factor shrinks the rate.

Try this

Q1. Differentiate $f(x) = e^{4x}$ and $g(x) = \ln(2x + 5)$. [2+2 marks]

• Cue. $f'(x) = 4e^{4x}$; $g'(x) = \frac{2}{2x + 5}$.

Q2. Find the gradient of $y = xe^{x}$ at $x = 1$. [3 marks]

• Cue. Product rule: $y' = e^x + xe^x = e^x(1 + x)$; at $x = 1$, $y' = 2e$.

Q3. A radioactive sample has mass $M = 50e^{-0.1t}$ grams. (a) Find $\frac{dM}{dt}$. (b) Find the rate of decay at $t = 5$. [2+2 marks]

• Cue. (a) $\frac{dM}{dt} = -5e^{-0.1t}$. (b) At $t = 5$, $-5e^{-0.5} = -3.03$ g per unit time.