Logarithmic and exponential calculus: derivatives and integrals of e^x and ln(x)

Use $\frac{d}{dx} \ln f(x) = \frac{f'(x)}{f(x)}$ with $f(x) = 3 x^2 + 1$.

$f'(x) = 6 x$.

$\frac{dy}{dx} = \frac{6 x}{3 x^2 + 1}$.

Markers reward identifying the inside function, stating the standard rule, and writing a clean unsimplified-but-correct quotient.

Notice that the numerator is exactly the derivative of the denominator. Use $\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C$.

With $f(x) = x^2 + 5$, $f'(x) = 2 x$.

$\int \frac{2 x}{x^2 + 5} \, dx = \ln (x^2 + 5) + C$.

No absolute value bars are needed since $x^2 + 5 > 0$ for all real $x$.

Markers reward recognising the reverse chain rule pattern and stating the integral with the correct constant of integration.

Differentiate with the chain rule. For $e^{f(x)}$ the derivative is $f'(x) e^{f(x)}$, and here $f(x) = 4x$ so $f'(x) = 4$:

Reverse the process for the integral. Antidifferentiating $e^{4x}$ undoes the multiply-by-$4$, so we divide by $4$:

Marker's note: one mark for $\frac{dy}{dx} = 4 e^{4x}$ with the chain-rule factor of $4$ shown, one for $\frac{e^{4x}}{4} + C$ with the constant of integration. Writing $4 e^{4x}$ for the integral (forgetting to divide) is the usual slip.

Apply the log chain rule. For $\ln f(x)$ the derivative is $\dfrac{f'(x)}{f(x)}$. Here $f(x) = 5x^2 - 3$, so $f'(x) = 10x$:

Find where the derivative is undefined. The quotient is undefined where the denominator is zero:

Marker's note: one mark for identifying $f'(x) = 10x$, one for the correct quotient $\frac{10x}{5x^2 - 3}$, one for $x = \pm\sqrt{3/5}$ from setting the denominator to zero. A candidate who differentiates but never solves the denominator loses the last mark.

Check whether the top is the derivative of the bottom. With $f(x) = x^3 + 2$, $f'(x) = 3x^2$. The numerator $6x^2 = 2 \times 3x^2 = 2 f'(x)$, so factor the $2$ out:

Apply the reverse chain rule $\int \frac{f'}{f} = \ln|f|$.

Marker's note: one mark for spotting the reverse-chain-rule pattern and factoring the constant $2$, one for $2\ln|x^3 + 2| + C$ with bars and the constant. Leaving the $2$ inside or dropping it entirely is the common error.

Set up the quotient rule. Let $u = \ln x$ and $v = x^2$, so $u' = \dfrac{1}{x}$ and $v' = 2x$. The quotient rule is $\dfrac{u'v - uv'}{v^2}$:

Simplify by cancelling a factor of $x$.

Marker's note: one mark for correct $u, v, u', v'$, one for the quotient-rule substitution, one for the simplified $\frac{1 - 2\ln x}{x^3}$. Forgetting to differentiate $\ln x$ as $\frac{1}{x}$, or failing to cancel the common $x$, are the usual marks lost.

Part (a): differentiate with the product rule. Let $u = x^2$ and $v = \ln x$, so $u' = 2x$ and $v' = \dfrac{1}{x}$:

Part (b): solve $\frac{dy}{dx} = 0$. Since $x > 0$, the factor $x$ is never zero, so

The $y$-coordinate is

so the stationary point is $\left(e^{-1/2}, -\dfrac{1}{2e}\right)$.

Classify it. Differentiate $\frac{dy}{dx} = 2x\ln x + x$ again:

At $x = e^{-1/2}$, $\ln x = -\tfrac{1}{2}$, so $\frac{d^2y}{dx^2} = 2\left(-\tfrac{1}{2}\right) + 3 = 2 > 0$, giving a minimum.

Marker's note: one mark for the product-rule working reaching $x(2\ln x + 1)$ in (a); one for $x = e^{-1/2}$, one for the exact $y = -\frac{1}{2e}$, and one for the correct classification via $y'' > 0$ in (b). Evaluating the messy logs numerically without exact forms, or skipping the nature test, loses marks.

Part (a): read the rate graph. The growth rate $\frac{dP}{dt}$ is positive for all $t$, so the population is always increasing. Because the rate curve is decreasing towards zero, the population rises quickly at first and then more and more slowly, levelling off towards a ceiling.

Part (b): integrate the rate to recover $P$. Antidifferentiate $\frac{dP}{dt} = 3e^{-0.5t}$ using $\int e^{kt}\,dt = \dfrac{e^{kt}}{k}$ with $k = -0.5$:

Use the initial condition $P = 2$ at $t = 0$ (note $e^0 = 1$):

so

Part (c): take the limit. As $t \to \infty$, $e^{-0.5t} \to 0$, so

The colony approaches a limiting population of $8$ thousand bacteria.

Marker's note: one mark for reading the rate graph as always-positive-but-decreasing in (a); two for the integration reaching $-6e^{-0.5t} + C$ and applying $P(0) = 2$ to get $C = 8$ in (b); one for the correct model $P = 8 - 6e^{-0.5t}$; one for the limiting value $8$ (thousand) in (c). Dividing by $-0.5$ incorrectly, or dropping the constant of integration before using the initial condition, are the marks most often lost.

Part (a): find the point and gradient. At $x = e$, $y = \ln e = 1$, so the point of contact is $(e, 1)$. Differentiating, $\dfrac{dy}{dx} = \dfrac{1}{x}$, so at $x = e$ the gradient is

The tangent through $(e, 1)$ with gradient $\frac{1}{e}$ is

So the tangent is $y = \dfrac{x}{e}$.

Part (b): find where the tangent meets the $x$-axis. Set $y = 0$:

so the tangent passes through the origin. The triangle has vertices at the origin $(0, 0)$, the foot of the vertical line at $(e, 0)$, and the point $(e, 1)$ on the line $x = e$.

Compute the area. The base runs along the $x$-axis from $x = 0$ to $x = e$, so the base length is $e$; the height is the vertical rise to $(e, 1)$, which is $1$:

Marker's note: one mark for the point $(e, 1)$ and gradient $\frac{1}{e}$, one for the tangent equation $y = \frac{x}{e}$, one for finding the $x$-intercept at the origin, one for identifying base $e$ and height $1$, one for the shown area $\frac{e}{2}$. A candidate who mis-simplifies the tangent to a non-zero intercept cannot reach the clean $\frac{e}{2}$, so careful algebra in (a) is what unlocks (b).