Find derivatives and integrals of $e^x$ and $\ln x$, including composed forms, and apply them to modelling problems

What this dot point is asking

NESA wants you to differentiate and integrate $e^x$, $\ln x$ and their composed forms with confidence. These functions appear in nearly every applied calculus problem: growth and decay, finance, biology and physics. Mastery here unlocks the modelling questions.

The answer

Standard derivatives

The exponential function is the unique function (up to a constant multiple) that is its own derivative. The logarithm is the inverse of $e^x$.

Composed derivatives (chain rule)

The second result is the basis of "logarithmic differentiation" and of the reverse-chain-rule integral $\int \frac{f'}{f} = \ln |f|$.

Standard integrals

The absolute value handles negative $x$, since $\ln$ is only defined for positive numbers. In a definite integral, the interval must not include $x = 0$.

Composed integrals

For a linear inside argument:

For the reverse chain rule on a logarithm:

If the numerator is a constant multiple of $f'(x)$, factor that constant out first.

Non-$e$ bases

For $a^x$ with $a > 0$, write $a^x = e^{x \ln a}$. Then

For $\log_a x$, use $\log_a x = \frac{\ln x}{\ln a}$, giving $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$.

Worked examples

Chain rule with IMATH_24

Differentiate $y = e^{\sin x}$.

$\frac{dy}{dx} = \cos x \cdot e^{\sin x}$.

Product rule with IMATH_27

Differentiate $y = x \ln x$.

$u = x$, $v = \ln x$. $u' = 1$, $v' = \frac{1}{x}$.

$\frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.

This derivative is zero at $x = e^{-1}$, which marks the minimum of $x \ln x$ for $x > 0$.

Quotient with IMATH_37

Differentiate $y = \frac{e^x}{x}$.

Quotient rule: $\frac{dy}{dx} = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x (x - 1)}{x^2}$.

Reverse chain rule integral

Evaluate $\int \frac{3 x^2}{x^3 + 4} \, dx$.

The numerator is $f'(x)$ for $f(x) = x^3 + 4$, so the integral is $\ln |x^3 + 4| + C$.

Definite integral

Evaluate $\int_1^e \frac{1}{x} \, dx$.

$\int_1^e \frac{1}{x} \, dx = [\ln x]_1^e = \ln e - \ln 1 = 1 - 0 = 1$.

This is the classic geometric definition of $e$: the number for which the area under $y = 1/x$ from $1$ to $e$ equals one.

Common traps

**$\int \frac{1}{x} \, dx \neq \frac{x^0}{0}$.** The power rule for integration excludes $n = -1$. The antiderivative of $\frac{1}{x}$ is $\ln |x|$, not a power.

Forgetting absolute value bars. Write $\int \frac{1}{x} \, dx = \ln |x| + C$. The bars are essential for intervals where $x < 0$.

Treating $e^{2 x}$ like $2 e^x$. They are different functions. $\frac{d}{dx}(e^{2 x}) = 2 e^{2 x}$ by the chain rule.

Dropping the chain rule on $\ln$. $\frac{d}{dx}(\ln(3 x)) = \frac{3}{3 x} = \frac{1}{x}$, but $\frac{d}{dx}(\ln(x^2)) = \frac{2 x}{x^2} = \frac{2}{x}$ (which also equals $\frac{d}{dx}(2 \ln x)$).

Splitting $\ln(a + b)$. $\ln(a + b)$ does not equal $\ln a + \ln b$. Only $\ln(a b) = \ln a + \ln b$ and $\ln(a/b) = \ln a - \ln b$.

In one sentence

The exponential function is its own derivative, the natural logarithm differentiates to $1/x$, and chain-rule and reverse-chain-rule patterns handle every composed form on the HSC Maths Advanced course.