Question: Differentiate $y = (x-1)\sqrt{4x+1}$ with respect to $x$ and simplify.

Step 1. Identify the structure: product of $u = x-1$ and $v = \sqrt{4x+1} = (4x+1)^{1/2}$.

Step 2. Differentiate each factor:

• $\dfrac{du}{dx} = 1$
• $\dfrac{dv}{dx} = \dfrac{1}{2}(4x+1)^{-1/2}(4) = \dfrac{2}{\sqrt{4x+1}}$ (chain rule)

Step 3. Apply product rule $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$:

$$\dfrac{dy}{dx} = (x-1) \cdot \dfrac{2}{\sqrt{4x+1}} + \sqrt{4x+1} \cdot 1$$

Step 4. Combine over common denominator $\sqrt{4x+1}$:

$$\dfrac{dy}{dx} = \dfrac{2(x-1) + (4x+1)}{\sqrt{4x+1}} = \dfrac{6x - 1}{\sqrt{4x+1}}$$

Tip: Always identify the structure first — product, quotient, or chain — before differentiating. For $\sqrt{f(x)}$, rewrite as $f(x)^{1/2}$ to keep the chain rule visible.

Question: Show that $\dfrac{d}{dx}\left[x e^{2x+1}\right] = e^{2x+1}(1 + 2x)$. Hence find $\displaystyle\int x e^{2x+1}\,dx$.

Step 1. Apply product rule to $x \cdot e^{2x+1}$:

$$\dfrac{d}{dx}\left[x e^{2x+1}\right] = 1 \cdot e^{2x+1} + x \cdot 2e^{2x+1} = e^{2x+1}(1 + 2x) \;\checkmark$$

Step 2. Integrate both sides (reverse the derivative):

$$\int e^{2x+1}(1 + 2x)\,dx = x e^{2x+1} + C$$

Step 3. Split the integrand into the part we want and the part we know:

$$\int e^{2x+1}\,dx + 2\int x e^{2x+1}\,dx = x e^{2x+1} + C$$

We know $\int e^{2x+1}\,dx = \dfrac{1}{2}e^{2x+1}$. Substitute:

$$\dfrac{1}{2}e^{2x+1} + 2\int x e^{2x+1}\,dx = x e^{2x+1} + C$$

Step 4. Solve for the wanted integral:

$$2\int x e^{2x+1}\,dx = x e^{2x+1} - \dfrac{1}{2}e^{2x+1} + C$$ $$\int x e^{2x+1}\,dx = \dfrac{1}{2}x e^{2x+1} - \dfrac{1}{4}e^{2x+1} + C$$

Tip: "Show ... hence find" is AM's version of integration by parts. The first part is the antidote — use it directly.

Question: The curve $y = \dfrac{e^{2x}(3x+2)}{5}$ has a stationary point. Find its coordinates and verify it is a minimum.

Step 1. Differentiate (product rule on $e^{2x}$ and $3x+2$):

$$\dfrac{dy}{dx} = \dfrac{1}{5}\left[2e^{2x}(3x+2) + 3e^{2x}\right] = \dfrac{e^{2x}}{5}(6x + 4 + 3) = \dfrac{e^{2x}(6x + 7)}{5}$$

Step 2. Set $\dfrac{dy}{dx} = 0$. Since $e^{2x} > 0$:

$$6x + 7 = 0 \Rightarrow x = -\dfrac{7}{6}$$

At $x = -\tfrac{7}{6}$: $y = \dfrac{e^{-7/3}\left(3 \cdot -\tfrac{7}{6} + 2\right)}{5} = \dfrac{e^{-7/3}\left(-\tfrac{1}{2}\right)}{5} = -\dfrac{e^{-7/3}}{10}$.

Step 3. Second derivative test:

$$\dfrac{d^2y}{dx^2} = \dfrac{1}{5}\left[2e^{2x}(6x+7) + 6e^{2x}\right] = \dfrac{e^{2x}(12x + 20)}{5}$$

At $x = -\tfrac{7}{6}$: $\dfrac{d^2y}{dx^2} = \dfrac{e^{-7/3}(-14 + 20)}{5} = \dfrac{6e^{-7/3}}{5} > 0$ $\Rightarrow$ minimum. $\checkmark$

Answer: Min at $\left(-\dfrac{7}{6}, -\dfrac{e^{-7/3}}{10}\right)$.

Tip: For $f \cdot g$ stationary points, the algebra often simplifies dramatically — group the common $e^{kx}$ factor and solve the bracket alone.

Question: The curve $y = x^3 - 3x^2 + 3x + 2$ has exactly one stationary point. Find its coordinates and determine its nature.

Step 1. Differentiate and set $\dfrac{dy}{dx} = 0$:

$$\frac{dy}{dx} = 3x^2 - 6x + 3 = 3(x - 1)^2 = 0 \implies x = 1$$

Then $y(1) = 1 - 3 + 3 + 2 = 3$, so the stationary point is $(1, 3)$.

Step 2. Try the second derivative:

$$\frac{d^2y}{dx^2} = 6x - 6 = 0 \;\text{ at }\; x = 1$$

The test is inconclusive — it tells us nothing here, so fall back to the sign of $\dfrac{dy}{dx}$.

Step 3. First-derivative sign test either side of $x = 1$:

$$x = 0.9:\;\; \frac{dy}{dx} = 3(-0.1)^2 = +0.03 \qquad x = 1:\;\; 0 \qquad x = 1.1:\;\; \frac{dy}{dx} = 3(0.1)^2 = +0.03$$

The gradient goes $+,\; 0,\; +$ — positive on both sides: the curve flattens momentarily and keeps rising, so this is neither a maximum nor a minimum.

$(1, 3)$ is a stationary point of inflexion.

⚠ Watch out: $\dfrac{d^2y}{dx^2} = 0$ does not prove an inflexion — $y = x^4$ has $y'' = 0$ at $x = 0$, yet that point is a minimum. Whenever the second derivative comes out zero, you must do the first-derivative sign test; quoting "$y'' = 0$, hence inflexion" loses the mark.

Question: Find the equation of the tangent to $y = (x-3)\sqrt{2x+3}$ at $x = 11$.

Step 1. Compute $y$ at $x=11$: $y = (11-3)\sqrt{25} = 8 \cdot 5 = 40$. Point: $(11, 40)$.

Step 2. Differentiate by product rule. Let $u = x-3$, $v = (2x+3)^{1/2}$:

• $u' = 1$, $v' = \dfrac{1}{2}(2x+3)^{-1/2} \cdot 2 = \dfrac{1}{\sqrt{2x+3}}$

$$\dfrac{dy}{dx} = \sqrt{2x+3} + \dfrac{x-3}{\sqrt{2x+3}} = \dfrac{(2x+3) + (x-3)}{\sqrt{2x+3}} = \dfrac{3x}{\sqrt{2x+3}}$$

Step 3. Evaluate at $x = 11$: $\dfrac{dy}{dx}\bigg|_{x=11} = \dfrac{33}{\sqrt{25}} = \dfrac{33}{5}$.

Step 4. Tangent equation via $y - y_1 = m(x - x_1)$:

$$y - 40 = \dfrac{33}{5}(x - 11) \;\Longrightarrow\; y = \dfrac{33}{5}x - \dfrac{363}{5} + 40 = \dfrac{33}{5}x - \dfrac{163}{5}$$

Tip: For the normal, replace $m$ with $-\dfrac{1}{m}$. Always confirm the point lies on the curve before plugging into the line equation.

Question: Explain why $y = 5x - \dfrac{1}{2}e^{-2x+1}$ is always increasing for all real $x$.

Step 1. Differentiate:

$$\dfrac{dy}{dx} = 5 - \dfrac{1}{2} \cdot (-2)e^{-2x+1} = 5 + e^{-2x+1}$$

Step 2. Analyse the sign. For all real $x$:

• $5 > 0$ (constant)
• $e^{-2x+1} > 0$ (exponential is always positive)

So $\dfrac{dy}{dx} = 5 + e^{-2x+1} > 5 > 0$ for all $x$.

Step 3. Conclude: since $\dfrac{dy}{dx} > 0$ everywhere, $y$ is always increasing. $\checkmark$

Tip: "Always increasing" arguments rely on showing $\dfrac{dy}{dx}$ never reaches zero or goes negative. Look for: positive constant + positive exponential, perfect square + positive constant, or $\dfrac{1}{(\text{positive})^2}$-style expressions where the sign is obvious without solving an inequality.

Question: Given $y = 3e^{2x} + 2e^{-x}$, find constants $A$ and $B$ such that $\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - y = A e^{2x} + B e^{-x}$.

Step 1. First derivative:

$$\dfrac{dy}{dx} = 6e^{2x} - 2e^{-x}$$

Step 2. Second derivative:

$$\dfrac{d^2y}{dx^2} = 12e^{2x} + 2e^{-x}$$

Step 3. Substitute into the LHS:

$$\dfrac{d^2y}{dx^2} + \dfrac{dy}{dx} - y = (12e^{2x} + 2e^{-x}) + (6e^{2x} - 2e^{-x}) - (3e^{2x} + 2e^{-x})$$ $$= (12 + 6 - 3)e^{2x} + (2 - 2 - 2)e^{-x} = 15 e^{2x} - 2 e^{-x}$$

Step 4. Compare with $A e^{2x} + B e^{-x}$: $A = 15$, $B = -2$.

Tip: ODE identity questions test bookkeeping. Compute each derivative cleanly, line up coefficients of $e^{2x}$ in one column and $e^{-x}$ in another, sum and read off.

Question: The surface area $A$ of a cube increases at $0.2$ cm²/s. Find the rate of increase of its volume $V$ when the side length is $1$ cm.

Step 1. Set up the formulas: $A = 6s^2$, $V = s^3$ where $s$ = side length.

Step 2. Differentiate w.r.t. $s$:

$$\dfrac{dA}{ds} = 12s, \qquad \dfrac{dV}{ds} = 3s^2$$

Step 3. Chain rule on $A$: $\dfrac{dA}{dt} = \dfrac{dA}{ds} \cdot \dfrac{ds}{dt}$, so $0.2 = 12s \cdot \dfrac{ds}{dt}$.

At $s = 1$: $\dfrac{ds}{dt} = \dfrac{0.2}{12} = \dfrac{1}{60}$ cm/s.

Step 4. Chain rule on $V$: $\dfrac{dV}{dt} = \dfrac{dV}{ds} \cdot \dfrac{ds}{dt} = 3s^2 \cdot \dfrac{1}{60}$.

At $s = 1$: $\dfrac{dV}{dt} = 3(1)^2 \cdot \dfrac{1}{60} = \dfrac{1}{20} = 0.05$ cm³/s.

Tip: Related rates always chain through a common variable. Map the chain on paper: $A \to s \to V$. Differentiate the formulas, plug numbers in last.

Question: A curve has $f'(x) = 9e^{-3x}$ and passes through $(0, -9)$ with a stationary point at $x = -\tfrac{1}{3}$. Find $f(x)$.

Wait — re-read. Two conditions needed: a curve point AND a stationary point. Stationary means $f'$ = 0 there, which contradicts $f'(x) = 9e^{-3x}$ (never zero). So this question must mean a higher-order curve. Re-state as: $f''(x) = 9e^{-3x}$, point $(0, -9)$, stationary at $x = -\tfrac{1}{3}$.

Step 1. First antiderivative:

$$f'(x) = \int 9e^{-3x}\,dx = -3e^{-3x} + C_1$$

Step 2. Stationary at $x = -\tfrac{1}{3}$: $f'(-\tfrac{1}{3}) = 0$, so $-3e^{1} + C_1 = 0 \Rightarrow C_1 = 3e$.

So $f'(x) = -3e^{-3x} + 3e$.

Step 3. Second antiderivative:

$$f(x) = \int (-3e^{-3x} + 3e)\,dx = e^{-3x} + 3ex + C_2$$

Step 4. Use $(0, -9)$: $1 + 0 + C_2 = -9 \Rightarrow C_2 = -10$.

$$\boxed{f(x) = e^{-3x} + 3ex - 10}$$

Tip: Anti-derivative construction: count conditions vs. constants. One constant per antiderivative step. Use stationary-point conditions on $f'$ first, then curve-point conditions on $f$.