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

Why $e$ is the natural base

The number $e \approx 2.71828$ is special precisely because $\frac{d}{dx}(e^x) = e^x$: the exponential to base $e$ is the only exponential function whose gradient at every point equals its own height. For any other base $a$, the derivative picks up a factor of $\ln a$, which is why all calculus is done in base $e$. The natural logarithm $\ln x = \log_e x$ is its inverse, and the pair $e^x$ and $\ln x$ undo one another: $e^{\ln x} = x$ for $x > 0$ and $\ln(e^x) = x$ for all $x$. These identities let you solve exponential equations by taking logs and logarithmic equations by exponentiating.

Why $e^x$ is its own derivative

The defining property of $e^x$ is that its gradient at every point equals its height there. That is what makes $\frac{d}{dx}e^x = e^x$, and seeing it makes the rule unforgettable.

Stage 1, plot $y = e^x$. Draw the exponential curve: it passes through $(0, 1)$, hugs the $x$-axis as $x$ decreases, and rises ever more steeply as $x$ increases.

Stage 2, the tangent at $(0, 1)$. The gradient of the tangent at $(0, 1)$ is $1$, which is exactly the height of the curve there, also $1$. Gradient equals height at this point.

Stage 3, the tangent at $(1, e)$. Move along to $(1, e)$, where the height is $e \approx 2.72$. The tangent there has gradient $e$ as well. Again the gradient matches the height.

Stage 4, gradient equals height everywhere. At every point the tangent gradient equals the $y$-value, so the gradient function is the curve itself: $\frac{d}{dx}e^x = e^x$. No other base has this property, which is precisely why all calculus is done in base $e$.

Log laws you will use inside calculus

Before differentiating or integrating, simplify with the log laws: $\ln(ab) = \ln a + \ln b$, $\ln(a/b) = \ln a - \ln b$, and $\ln(a^n) = n \ln a$. Rewriting $\ln(x^3)$ as $3 \ln x$ before differentiating turns a chain-rule problem into a one-line answer, $\frac{3}{x}$. Splitting a product or quotient inside a logarithm into a sum or difference of logs is often the fastest route through a derivative.

Curve features from the calculus

Differentiation reveals the shape of these graphs. For $y = x e^{-x}$, the derivative $y' = e^{-x}(1 - x)$ is zero at $x = 1$, a maximum; the curve rises then decays towards the $x$-axis. For $y = \ln x$, $y' = \frac{1}{x} > 0$ for all $x > 0$, so $\ln x$ is always increasing, while $y'' = -\frac{1}{x^2} < 0$ shows it is always concave down. Reading derivative information back into the graph is a standard HSC application.

How exam questions ask about exponential and logarithmic calculus

• "Differentiate $e^{f(x)}$ or $\ln f(x)$." Apply the composed rule; show $f'(x)$. For a product or quotient with $e^x$ or $\ln x$, combine with the product or quotient rule.
• "Find $\int e^{kx}\,dx$." Divide by $k$: $\int e^{kx}\,dx = \frac{e^{kx}}{k} + C$.
• "Find $\int \frac{f'(x)}{f(x)}\,dx$" or an integral with the derivative on top. Recognise the reverse chain rule and write $\ln|f(x)| + C$; factor out any constant on the numerator first.
• "Evaluate $\int_a^b \frac{1}{x}\,dx$." The antiderivative is $\ln|x|$; check the interval avoids $x = 0$, then substitute the limits.
• "Find the stationary point / maximum of $y = x e^{-x}$" (or similar). Differentiate (product rule), set $=0$, and classify - exponential and log models are common optimisation contexts.
• "Differentiate $a^x$ or $\log_a x$" for a non-$e$ base. Convert with $a^x = e^{x\ln a}$ and $\log_a x = \frac{\ln x}{\ln a}$, giving the extra factor of $\ln a$.