Differentiation rules for HSC Maths Advanced: power, chain, product, quotient, exp, log, trig

This is a composition, so use the chain rule. Let $u = 3x^2 + 1$, so $y = u^5$.

$\frac{dy}{du} = 5u^4$ and $\frac{du}{dx} = 6x$.

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 5(3x^2 + 1)^4 \cdot 6x = 30x(3x^2 + 1)^4$.

Markers reward identifying the inner and outer functions, applying the chain rule cleanly, and presenting the final answer in factored form.

Use the product rule with $u = x^2$ and $v = \ln x$.

$u' = 2x$ and $v' = \frac{1}{x}$.

$f'(x) = u' v + u v' = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x$.

Factor: $f'(x) = x(2 \ln x + 1)$.

Markers expect explicit labelling of $u$, $v$ and their derivatives, correct application of the rule, and a tidy final answer.

Recognise a composition, so use the chain rule. Let $u = 2x + 5$, giving $y = u^7$.

Multiply the two rates.

Marker's note: one mark for the chain-rule setup (identifying $u = 2x + 5$ and differentiating the outer power), one for the correct final answer $14(2x + 5)^6$. Forgetting the inner factor of $2$ and writing $7(2x + 5)^6$ is the standard slip that loses the second mark.

Name the two factors. Let $u = x^3$ and $v = e^x$, so

Apply $f' = u'v + uv'$.

Marker's note: one mark for the correct product-rule expansion $3x^2 e^x + x^3 e^x$, one for a tidy factored answer $x^2 e^x(3 + x)$. Markers award the method mark for correct $u, v, u', v'$ even if the factoring slips.

Set up the quotient rule. Let $u = 2x - 1$ and $v = x^2 + 3$, so

Apply $\dfrac{u'v - uv'}{v^2}$, keeping the order in the numerator.

Expand and simplify the numerator.

so

Marker's note: one mark for the correct quotient-rule structure with $u'v$ before $uv'$, one for expanding the numerator correctly, one for the simplified final answer. Reversing the numerator to $uv' - u'v$ flips every sign and loses the last two marks.

Find the point of contact. At $x = 1$, $y = 1 \cdot \ln 1 = 1 \cdot 0 = 0$, so the tangent touches at $(1, 0)$.

Differentiate with the product rule. Let $u = x$ and $v = \ln x$, so $u' = 1$ and $v' = \dfrac{1}{x}$:

Evaluate the gradient at $x = 1$.

Write the line through $(1, 0)$ with gradient $1$.

Marker's note: one mark for the product-rule derivative $\ln x + 1$, one for evaluating the gradient as $1$ at $x = 1$, one for the correct tangent equation $y = x - 1$. Using $\ln 1 = 1$ instead of $0$ is the usual error.

Apply the quotient rule. Let $u = \cos x$ and $v = x^2$, so $u' = -\sin x$ and $v' = 2x$:

Cancel a factor of $x$.

Substitute $x = \pi$ (radians). Here $\sin \pi = 0$ and $\cos \pi = -1$:

Marker's note: one mark for the quotient-rule setup, one for the simplified derivative, one for the exact value $\dfrac{2}{\pi^3}$ using $\sin \pi = 0$ and $\cos \pi = -1$. Working in degrees, or slipping the sign of $\cos \pi$, breaks the final mark.

Velocity is the derivative of displacement. Differentiate $x = t^2 e^{-t}$ with the product rule, letting $u = t^2$ and $v = e^{-t}$, so $u' = 2t$ and $v' = -e^{-t}$:

Solve $v = 0$. Since $e^{-t} > 0$ for all $t$, the velocity is zero only when

Interpret each time using the described graph. At $t = 0$ the particle starts from rest at the origin (the velocity graph begins at the origin). At $t = 2$ the velocity graph is at its single peak turning point, so the particle is momentarily at rest at the far end of its outward motion before slowing displacement growth; this is where displacement stops increasing as fast and the velocity has returned to zero.

Marker's note: one mark for the product-rule velocity $v = t\,e^{-t}(2 - t)$, one for solving $v = 0$ to $t = 0$ and $t = 2$ (using $e^{-t} \neq 0$), one for reading the graph to say the particle is at rest at both times. Cancelling or ignoring the $e^{-t}$ factor as if it could be zero is the trap.

Part (a): differentiate with the quotient rule. Let $u = e^{2x}$ and $v = x$. By the chain rule $u' = 2e^{2x}$, and $v' = 1$:

Factor $e^{2x}$ out of the numerator:

as required.

Part (b): locate the stationary point. Set $\dfrac{dy}{dx} = 0$. Since $e^{2x} > 0$ and $x^2 > 0$ for $x > 0$, the only zero comes from

The $y$-coordinate is

so the stationary point is $\left(\tfrac{1}{2}, 2e\right)$.

Justify the minimum with a sign test. The factors $e^{2x}$ and $x^2$ are always positive for $x > 0$, so the sign of $\dfrac{dy}{dx}$ follows the sign of $(2x - 1)$. For $x < \tfrac{1}{2}$, $(2x - 1) < 0$ so $\dfrac{dy}{dx} < 0$ (falling); for $x > \tfrac{1}{2}$, $(2x - 1) > 0$ so $\dfrac{dy}{dx} > 0$ (rising). The gradient changes from negative to positive, so $\left(\tfrac{1}{2}, 2e\right)$ is a minimum.

Part (c): confirm the value. From part (b), $y = \dfrac{e^{1}}{1/2} = 2e$, which is the required result.

Marker's note: one mark for the quotient-and-chain-rule derivative reaching the shown form in (a); one for solving $\dfrac{dy}{dx} = 0$ to $x = \tfrac{1}{2}$; one for the sign test either side confirming a minimum; one for the coordinates $\left(\tfrac{1}{2}, 2e\right)$; one for showing $y = 2e$ cleanly in (c). Forgetting the chain-rule factor of $2$ on $e^{2x}$ derails part (a) and every mark after it.

Part (a): differentiate twice. First derivative by the product rule with $u = e^{-x}$, $v = \cos x$, so $u' = -e^{-x}$, $v' = -\sin x$:

Differentiate again, using the product rule on $-e^{-x}(\cos x + \sin x)$ with $u = -e^{-x}$ (so $u' = e^{-x}$) and $v = \cos x + \sin x$ (so $v' = -\sin x + \cos x$):

Expand inside:

as required.

Part (b): possible inflections where $f''(x) = 0$. Since $e^{-x} > 0$ for all $x$, we need $\sin x = 0$. On $0 \le x \le 2\pi$ (radians),

These are the $x$-values where the concavity may change, so the possible points of inflection are at $x = 0$, $x = \pi$ and $x = 2\pi$.

Marker's note: one mark for the first derivative $-e^{-x}(\cos x + \sin x)$, one for reaching $f''(x) = 2e^{-x}\sin x$, one for setting $\sin x = 0$ using $e^{-x} \neq 0$, one for all three solutions in the given domain. Working in degrees or dropping a solution at an endpoint costs the final mark.