A-Level Mathematics – Chapter 6: Differentiation
Differentiation is the mathematical study of rates of change and the gradients of curves. It underpins curve sketching, optimisation, mechanics, and an enormous range of applied mathematics. This chapter develops the subject from the foundational limit definition, through standard derivatives and the chain, product, and quotient rules, to the Year 2 techniques of implicit and parametric differentiation, before arriving at stationary points and optimisation problems.
Specification Note
Content labelled Year 2 is A2-level only and is not assessed at AS. Sections 6.1, 6.2, and 6.6 are required for both AS and full A-Level.
Contents
6.1 First Principles
The derivative is defined as the limit of the gradient of a chord as the chord length tends to zero. This formal definition underpins all of differentiation, and being able to derive simple results from it is an explicit exam requirement.
Definition: The Derivative from First Principles
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$Also written as $\dfrac{dy}{dx}$, $\dfrac{d}{dx}[f(x)]$, or $\dot{y}$ (in mechanics).
Geometrically, $f'(x)$ gives the gradient of the tangent to $y = f(x)$ at the point $(x, f(x))$, and also the instantaneous rate of change of $y$ with respect to $x$.
Standard Limits Required
Two limits are needed for the first-principles derivation of trigonometric derivatives (both stated without proof at A-Level):
$$\lim_{h \to 0}\frac{\sin h}{h} = 1 \qquad \lim_{h \to 0}\frac{\cos h - 1}{h} = 0$$Example 6.1.1 — Differentiating $f(x) = x^2$ from First Principles
Use the definition to show that $\dfrac{d}{dx}(x^2) = 2x$.
1 Form the difference quotient: $\dfrac{(x+h)^2 - x^2}{h}$
2 Expand: $= \dfrac{x^2 + 2xh + h^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h$
3 Take the limit: $f'(x) = \lim_{h \to 0}(2x + h) = 2x$
Example 6.1.2 — Differentiating $f(x) = x^3$ from First Principles
Show that $\dfrac{d}{dx}(x^3) = 3x^2$.
1 $\dfrac{(x+h)^3 - x^3}{h}$
2 Expand $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$:
$= \dfrac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2$
3 $\lim_{h\to 0}(3x^2 + 3xh + h^2) = 3x^2$
Example 6.1.3 — Differentiating $\sin x$ from First Principles
Show that $\dfrac{d}{dx}(\sin x) = \cos x$.
1 $\dfrac{\sin(x+h) - \sin x}{h}$. Apply the compound angle formula: $\sin(x+h) = \sin x\cos h + \cos x\sin h$.
2 $= \dfrac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\dfrac{\cos h - 1}{h} + \cos x\cdot\dfrac{\sin h}{h}$
3 Apply the standard limits: $= \sin x \cdot 0 + \cos x \cdot 1 = \cos x$
Exam Tip
In first-principles questions, you must show the limit notation explicitly and take the limit at the final step. Writing $h \to 0$ once at the start and then substituting $h = 0$ will lose marks — keep the limit notation until the very last line.
6.2 Standard Derivatives
The following table of derivatives must be memorised. The power rule and the trig derivatives are given in the formulae booklet, but the $e^x$ and $\ln x$ results are not, so they should be learnt.
Table of Standard Derivatives
| $f(x)$ | $f'(x)$ |
|---|---|
| $x^n$ | $nx^{n-1}$ |
| $e^x$ | $e^x$ |
| $e^{ax}$ | $ae^{ax}$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
Sum/difference rule: $\dfrac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$
Constant multiple: $\dfrac{d}{dx}[kf(x)] = kf'(x)$
The figure below shows a function $f(x)$ alongside its derivative $f'(x)$. Observe how the derivative is zero where the original function has a horizontal tangent, positive where the function increases, and negative where it decreases.
Figure 6.1 — $f(x) = x^3 - 3x$ (blue) and its derivative $f'(x) = 3x^2 - 3$ (green). The derivative is zero at the turning points of $f$.
Example 6.2.1 — Differentiating a Polynomial (Rewriting First)
Find $\dfrac{dy}{dx}$ for $y = 4x^3 - \dfrac{6}{x^2} + 5\sqrt{x} - 7$.
1 Rewrite: $y = 4x^3 - 6x^{-2} + 5x^{1/2} - 7$
2 Differentiate term by term:
$\dfrac{dy}{dx} = 12x^2 + 12x^{-3} + \dfrac{5}{2}x^{-1/2} = 12x^2 + \dfrac{12}{x^3} + \dfrac{5}{2\sqrt{x}}$
Example 6.2.2 — Differentiating Trigonometric and Exponential Functions
Differentiate: (a) $y = 3\sin x - 2\cos x$ (b) $y = 4e^{3x} + \ln x$ (c) $y = 2\tan x - 5e^x$
a $\dfrac{dy}{dx} = 3\cos x + 2\sin x$
b $\dfrac{dy}{dx} = 12e^{3x} + \dfrac{1}{x}$
c $\dfrac{dy}{dx} = 2\sec^2 x - 5e^x$
Example 6.2.3 — Finding the Gradient at a Point
Find the gradient of $y = x^3 - 4x + 1$ at the point where $x = 2$.
1 $\dfrac{dy}{dx} = 3x^2 - 4$
2 At $x = 2$: $\dfrac{dy}{dx} = 12 - 4 = 8$
Exam Tip
Before differentiating, always rewrite the function in index notation: $\dfrac{1}{x^2} = x^{-2}$, $\sqrt[3]{x} = x^{1/3}$, $\dfrac{5}{\sqrt{x}} = 5x^{-1/2}$. Attempting to apply the power rule without rewriting is the most common error in this section.
6.3 Rules of Differentiation
The chain rule, product rule, and quotient rule extend differentiation to composite, product, and quotient functions respectively. These three rules are the workhorses of Year 1 and Year 2 calculus.
Chain Rule
For a composite function $y = f(g(x))$:
$$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) \qquad \text{equivalently,} \quad \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$$Mnemonic: differentiate the outer function, leave the inner unchanged, then multiply by the derivative of the inner.
Product Rule
For $y = u(x)\,v(x)$:
$$\frac{dy}{dx} = u'v + uv'$$Quotient Rule
For $y = \dfrac{u(x)}{v(x)}$:
$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$$Mnemonic: "bottom $d$(top) minus top $d$(bottom), over bottom squared."
Example 6.3.1 — Chain Rule: Power of a Function
Differentiate $y = (3x^2 + 1)^5$.
1 Let $u = 3x^2 + 1$, $y = u^5$.
2 $\dfrac{dy}{du} = 5u^4$ and $\dfrac{du}{dx} = 6x$.
3 $\dfrac{dy}{dx} = 5u^4 \cdot 6x = 30x(3x^2+1)^4$
Example 6.3.2 — Chain Rule: Logarithm and Trigonometry
Differentiate: (a) $y = \ln(4x^3 - 1)$ (b) $y = \sin(e^{2x})$
a Let $u = 4x^3 - 1$: $\dfrac{dy}{dx} = \dfrac{1}{u} \cdot 12x^2 = \dfrac{12x^2}{4x^3 - 1}$
b Let $u = e^{2x}$: $\dfrac{dy}{dx} = \cos(e^{2x}) \cdot 2e^{2x} = 2e^{2x}\cos(e^{2x})$
Example 6.3.3 — Product Rule
Differentiate $y = x^3 e^{2x}$.
1 $u = x^3$, $v = e^{2x}$; $u' = 3x^2$, $v' = 2e^{2x}$.
2 $\dfrac{dy}{dx} = 3x^2 e^{2x} + x^3 \cdot 2e^{2x} = x^2 e^{2x}(3 + 2x)$
Example 6.3.4 — Product Rule with $\ln x$
Differentiate $y = x^2\ln x$. Hence find $\dfrac{dy}{dx}$ at $x = e$.
1 $u = x^2$, $v = \ln x$; $u' = 2x$, $v' = \dfrac{1}{x}$.
2 $\dfrac{dy}{dx} = 2x\ln x + x^2 \cdot \dfrac{1}{x} = 2x\ln x + x$
3 At $x = e$: $\dfrac{dy}{dx} = 2e\ln e + e = 2e + e = 3e$
Example 6.3.5 — Quotient Rule
Differentiate $y = \dfrac{\sin x}{x^2 + 1}$.
1 $u = \sin x$, $v = x^2+1$; $u' = \cos x$, $v' = 2x$.
2 $\dfrac{dy}{dx} = \dfrac{\cos x(x^2+1) - \sin x \cdot 2x}{(x^2+1)^2} = \dfrac{(x^2+1)\cos x - 2x\sin x}{(x^2+1)^2}$
Example 6.3.6 — Chain Rule Inside Product Rule
Differentiate $y = e^{x^2}\sin(3x)$.
1 $u = e^{x^2}$, $v = \sin(3x)$; $u' = 2xe^{x^2}$, $v' = 3\cos(3x)$.
2 $\dfrac{dy}{dx} = 2xe^{x^2}\sin(3x) + e^{x^2} \cdot 3\cos(3x) = e^{x^2}[2x\sin(3x) + 3\cos(3x)]$
Exam Tip
Before applying the quotient rule, check whether the expression can be split and differentiated term by term instead — it is often faster. For instance, $\dfrac{x^3 + 2x}{x}$ should be split as $x^2 + 2$ rather than treated as a quotient.
6.4 Implicit Differentiation Year 2
When a curve is defined by an equation in both $x$ and $y$ that cannot easily be rearranged into $y = f(x)$, we differentiate implicitly with respect to $x$. The key idea is that $y$ is treated as a function of $x$, so every term involving $y$ acquires a factor of $\dfrac{dy}{dx}$ by the chain rule:
$$\frac{d}{dx}[f(y)] = f'(y)\,\frac{dy}{dx}$$Example 6.4.1 — A Circle
Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$ and hence find the gradient at the point $(3, 4)$.
1 Differentiate: $2x + 2y\dfrac{dy}{dx} = 0$
2 Solve: $\dfrac{dy}{dx} = -\dfrac{x}{y}$
3 At $(3,4)$: $\dfrac{dy}{dx} = -\dfrac{3}{4}$
Example 6.4.2 — Mixed Terms
Find the gradient of $x^3 + y^3 - 3xy = 7$ at the point $(2, 1)$.
1 Differentiate: $3x^2 + 3y^2\dfrac{dy}{dx} - 3\!\left(y + x\dfrac{dy}{dx}\right) = 0$
2 Rearrange: $(3y^2 - 3x)\dfrac{dy}{dx} = 3y - 3x^2$
3 $\dfrac{dy}{dx} = \dfrac{y - x^2}{y^2 - x}$
4 At $(2,1)$: $\dfrac{dy}{dx} = \dfrac{1 - 4}{1 - 2} = \dfrac{-3}{-1} = 3$
Example 6.4.3 — Finding the Tangent to an Implicit Curve
Find the equation of the tangent to $ye^x + x^2 = 3$ at the point $(0, 3)$.
1 Differentiate: $\dfrac{dy}{dx}e^x + ye^x + 2x = 0$
2 At $(0, 3)$: $\dfrac{dy}{dx}\cdot 1 + 3\cdot 1 + 0 = 0 \Rightarrow \dfrac{dy}{dx} = -3$
3 Tangent: $y - 3 = -3(x - 0) \Rightarrow y = -3x + 3$
Example 6.4.4 — Implicit Differentiation with $\ln$
Find $\dfrac{dy}{dx}$ for $\ln(xy) + x = 2y$.
1 Use log law: $\ln x + \ln y + x = 2y$
2 Differentiate: $\dfrac{1}{x} + \dfrac{1}{y}\dfrac{dy}{dx} + 1 = 2\dfrac{dy}{dx}$
3 Rearrange: $\left(2 - \dfrac{1}{y}\right)\dfrac{dy}{dx} = \dfrac{1}{x} + 1$
4 $\dfrac{dy}{dx} = \dfrac{\dfrac{1}{x} + 1}{2 - \dfrac{1}{y}} = \dfrac{y(1+x)}{x(2y-1)}$
Exam Tip
After differentiating implicitly, collect all $\dfrac{dy}{dx}$ terms on one side immediately — do not try to simplify while $\dfrac{dy}{dx}$ terms are scattered on both sides. Factorising cleanly is essential for full marks.
6.5 Parametric Differentiation Year 2
When a curve is given parametrically by $x = f(t)$ and $y = g(t)$, differentiate each with respect to $t$ and use the chain rule to find $\dfrac{dy}{dx}$:
Parametric Differentiation
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad \left(\text{provided } \frac{dx}{dt} \neq 0\right)$$The second derivative is found by:
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$$Example 6.5.1 — First Derivative
A curve is given by $x = t^2 + 1$, $y = 2t^3 - 3t$. Find $\dfrac{dy}{dx}$ and the gradient at $t = 1$.
1 $\dfrac{dx}{dt} = 2t$, $\dfrac{dy}{dt} = 6t^2 - 3$
2 $\dfrac{dy}{dx} = \dfrac{6t^2 - 3}{2t} = \dfrac{3(2t^2-1)}{2t}$
3 At $t=1$: $\dfrac{dy}{dx} = \dfrac{3(1)}{2} = \dfrac{3}{2}$
Example 6.5.2 — Second Derivative Parametrically
For the curve in Example 6.5.1, find $\dfrac{d^2y}{dx^2}$ at $t = 1$.
1 $\dfrac{dy}{dx} = \dfrac{3(2t^2-1)}{2t}$. Differentiate with respect to $t$ using the quotient rule:
$\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \dfrac{12t \cdot 2t - (6t^2-3)\cdot 2}{4t^2} = \dfrac{24t^2 - 12t^2 + 6}{4t^2} = \dfrac{12t^2+6}{4t^2} = \dfrac{3(2t^2+1)}{2t^2}$
2 $\dfrac{d^2y}{dx^2} = \dfrac{3(2t^2+1)/(2t^2)}{2t} = \dfrac{3(2t^2+1)}{4t^3}$
3 At $t=1$: $\dfrac{d^2y}{dx^2} = \dfrac{3\times3}{4} = \dfrac{9}{4}$
Example 6.5.3 — Ellipse: Tangent at a Point
A curve has parametric equations $x = 3\cos t$, $y = 2\sin t$. Find the equation of the tangent at $t = \dfrac{\pi}{4}$.
1 $\dfrac{dx}{dt} = -3\sin t$, $\dfrac{dy}{dt} = 2\cos t$.
2 $\dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2}{3}\cot t$
3 At $t = \pi/4$: $\cot(\pi/4) = 1$, so gradient $= -\dfrac{2}{3}$.
4 Point: $x = \dfrac{3}{\sqrt{2}}$, $y = \dfrac{2}{\sqrt{2}} = \sqrt{2}$.
5 Tangent: $y - \sqrt{2} = -\dfrac{2}{3}\!\left(x - \dfrac{3}{\sqrt{2}}\right) \Rightarrow y = -\dfrac{2}{3}x + \dfrac{\sqrt{2}}{1} + \sqrt{2} = -\dfrac{2}{3}x + 2\sqrt{2}$
Example 6.5.4 — Parametric Normal
Find the equation of the normal to $x = \ln(t+1)$, $y = t^2 - 2t$ at $t = 2$.
1 Point: $x = \ln 3$, $y = 4 - 4 = 0$.
2 $\dfrac{dx}{dt} = \dfrac{1}{t+1}$, $\dfrac{dy}{dt} = 2t-2$. So $\dfrac{dy}{dx} = (2t-2)(t+1)$.
3 At $t=2$: $\dfrac{dy}{dx} = (2)(3) = 6$. Normal gradient $= -\dfrac{1}{6}$.
4 Normal: $y - 0 = -\dfrac{1}{6}(x - \ln 3) \Rightarrow y = -\dfrac{1}{6}x + \dfrac{\ln 3}{6}$
Exam Tip
To find $\dfrac{d^2y}{dx^2}$ parametrically, do not differentiate $\dfrac{dy}{dx}$ with respect to $x$ — differentiate with respect to $t$ first, then divide by $\dfrac{dx}{dt}$. Forgetting to divide by $\dfrac{dx}{dt}$ is the most common error here.
6.6 Stationary Points and Optimisation
A stationary point is a point where $\dfrac{dy}{dx} = 0$. At such a point the tangent is horizontal. Stationary points fall into three categories: local maxima, local minima, and points of inflection (where the gradient does not change sign).
Classifying Stationary Points
At a stationary point $x = a$ where $f'(a) = 0$, examine the second derivative $f''(a)$:
- $f''(a) < 0$: local maximum
- $f''(a) > 0$: local minimum
- $f''(a) = 0$: inconclusive — use the first derivative test (examine sign of $f'$ on either side)
The figure below shows a cubic with its stationary points marked. The blue labels indicate the classification from the second derivative test.
Figure 6.2 — $y = x^3 - 3x$ (grey) with local maximum at $(-1, 2)$ (red) and local minimum at $(1, -2)$ (blue), found by setting $y' = 3x^2 - 3 = 0$.
Example 6.6.1 — Finding and Classifying Stationary Points
Find and classify all stationary points of $y = 2x^3 - 9x^2 + 12x - 4$.
1 $\dfrac{dy}{dx} = 6x^2 - 18x + 12 = 6(x-1)(x-2) = 0 \Rightarrow x = 1$ or $x = 2$
2 $\dfrac{d^2y}{dx^2} = 12x - 18$
3 At $x=1$: $y = 2-9+12-4 = 1$; $y'' = -6 < 0$ → local maximum at $(1, 1)$
4 At $x=2$: $y = 16-36+24-4 = 0$; $y'' = 6 > 0$ → local minimum at $(2, 0)$
Example 6.6.2 — Tangent and Normal
Find the equations of the tangent and normal to $y = x^3 - 3x$ at $x = 2$.
1 $y(2) = 8 - 6 = 2$. Point $(2, 2)$.
2 $y' = 3x^2 - 3$; at $x=2$: gradient $= 9$.
3 Tangent: $y - 2 = 9(x-2) \Rightarrow y = 9x - 16$
4 Normal gradient: $-\tfrac{1}{9}$. Normal: $y - 2 = -\tfrac{1}{9}(x-2) \Rightarrow x + 9y = 20$
Example 6.6.3 — Optimisation: Cylindrical Tin
A closed cylinder has volume $250\pi\,\text{cm}^3$. Find the dimensions that minimise the total surface area.
1 Let radius $= r$, height $= h$. Volume: $\pi r^2 h = 250\pi \Rightarrow h = \dfrac{250}{r^2}$.
2 $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{500\pi}{r}$
3 $\dfrac{dS}{dr} = 4\pi r - \dfrac{500\pi}{r^2} = 0 \Rightarrow r^3 = 125 \Rightarrow r = 5\,\text{cm}$
4 $h = \dfrac{250}{25} = 10\,\text{cm}$. $\dfrac{d^2S}{dr^2} > 0$ → minimum confirmed.
Example 6.6.4 — Connected Rates of Change
A spherical balloon is inflated so that its volume increases at $200\,\text{cm}^3\,\text{s}^{-1}$. Find the rate of increase of the radius when $r = 5\,\text{cm}$.
1 $V = \tfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dr} = 4\pi r^2$
2 Chain rule: $\dfrac{dr}{dt} = \dfrac{dV/dt}{dV/dr} = \dfrac{200}{4\pi r^2}$
3 At $r=5$: $\dfrac{dr}{dt} = \dfrac{200}{100\pi} = \dfrac{2}{\pi} \approx 0.637\,\text{cm\,s}^{-1}$
Example 6.6.5 — Optimisation: Rectangular Field
A farmer has 120 m of fencing and a straight wall as one side of a rectangle. Find the dimensions that maximise the enclosed area.
1 Let sides perpendicular to the wall each be $y$ m; side parallel to the wall is $x$ m. Then $x + 2y = 120 \Rightarrow x = 120 - 2y$.
2 $A = xy = (120-2y)y = 120y - 2y^2$
3 $\dfrac{dA}{dy} = 120 - 4y = 0 \Rightarrow y = 30\,\text{m}$; $x = 60\,\text{m}$
4 $\dfrac{d^2A}{dy^2} = -4 < 0$ → maximum. Area $= 1800\,\text{m}^2$.
Exam Tip
In optimisation, always show that you have verified the nature of the stationary point (maximum or minimum) — a one-line check with the second derivative or a sign diagram is required for full marks. Simply finding the stationary point is not sufficient.
Practice Problems
Problem 1
Differentiate $y = \dfrac{x^4 - 3x^2 + 7}{x}$ after first simplifying the expression.
Show Solution
Simplify: $y = x^3 - 3x + 7x^{-1}$
$\dfrac{dy}{dx} = 3x^2 - 3 - \dfrac{7}{x^2}$
Problem 2
Differentiate from first principles: $f(x) = 3x^2 + 2x$.
Show Solution
$\dfrac{f(x+h)-f(x)}{h} = \dfrac{3(x+h)^2 + 2(x+h) - 3x^2 - 2x}{h} = \dfrac{6xh + 3h^2 + 2h}{h} = 6x + 3h + 2$
$f'(x) = \lim_{h\to 0}(6x + 3h + 2) = 6x + 2$
Problem 3
Differentiate $y = \cos^4(3x)$ using the chain rule.
Show Solution
Let $u = \cos(3x)$, $y = u^4$.
$\dfrac{dy}{du} = 4u^3$, $\dfrac{du}{dx} = -3\sin(3x)$
$\dfrac{dy}{dx} = 4\cos^3(3x) \cdot (-3\sin(3x)) = -12\cos^3(3x)\sin(3x)$
Problem 4
Differentiate $y = \dfrac{e^x}{x^2 + 4}$ using the quotient rule.
Show Solution
$u = e^x$, $v = x^2+4$; $u' = e^x$, $v' = 2x$
$\dfrac{dy}{dx} = \dfrac{e^x(x^2+4) - 2xe^x}{(x^2+4)^2} = \dfrac{e^x(x^2 - 2x + 4)}{(x^2+4)^2}$
Problem 5 Year 2
Find $\dfrac{dy}{dx}$ by implicit differentiation for $e^{x+y} = xy + 1$.
Show Solution
Differentiate: $e^{x+y}\!\left(1 + \dfrac{dy}{dx}\right) = y + x\dfrac{dy}{dx}$
$(e^{x+y} - x)\dfrac{dy}{dx} = y - e^{x+y}$
$\dfrac{dy}{dx} = \dfrac{y - e^{x+y}}{e^{x+y} - x}$
Problem 6 Year 2
A curve has parametric equations $x = \ln(t+1)$, $y = t^2 - 2t$. Find the equation of the tangent at $t = 1$.
Show Solution
At $t=1$: $x = \ln 2$, $y = -1$.
$\dfrac{dx}{dt} = \dfrac{1}{t+1}$, $\dfrac{dy}{dt} = 2t-2$. So $\dfrac{dy}{dx} = (2t-2)(t+1)$.
At $t=1$: $\dfrac{dy}{dx} = (0)(2) = 0$. Tangent: $y = -1$.
Problem 7
Find and classify all stationary points of $y = x^4 - 8x^2 + 3$.
Show Solution
$y' = 4x^3 - 16x = 4x(x-2)(x+2) = 0 \Rightarrow x = 0, \pm 2$
$y'' = 12x^2 - 16$
$x=0$: $y'' = -16 < 0$ → local max at $(0, 3)$
$x=2$: $y'' = 32 > 0$ → local min at $(2, -13)$
$x=-2$: $y'' = 32 > 0$ → local min at $(-2, -13)$
Problem 8
A ladder of length 5 m leans against a vertical wall. Its foot slides away from the wall at $0.1\,\text{m\,s}^{-1}$. Find the rate at which the top slides down when the foot is 3 m from the wall.
Show Solution
Let $x$ = distance of foot from wall, $y$ = height of top. Then $x^2 + y^2 = 25$.
Differentiate: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt}$
When $x=3$: $y = 4$. $\dfrac{dy}{dt} = -\dfrac{3}{4}\times 0.1 = -0.075\,\text{m\,s}^{-1}$
The top slides down at $0.075\,\text{m\,s}^{-1}$.
Problem 9
Prove from first principles that $\dfrac{d}{dx}(\cos x) = -\sin x$.
Show Solution
$\lim_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h}$. Use $\cos(x+h) = \cos x\cos h - \sin x\sin h$:
$= \lim_{h\to 0}\left[\cos x\cdot\dfrac{\cos h - 1}{h} - \sin x\cdot\dfrac{\sin h}{h}\right]$
$= \cos x\cdot 0 - \sin x\cdot 1 = -\sin x$ QED
Problem 10 Year 2
The curve $C$ is defined by $x = 2t + 1$, $y = t^3 - 3t$. Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ and hence determine whether the point at $t = 2$ is a point of inflection, local maximum, or local minimum on the curve.
Show Solution
$\dfrac{dx}{dt} = 2$, $\dfrac{dy}{dt} = 3t^2 - 3$. So $\dfrac{dy}{dx} = \dfrac{3t^2-3}{2}$.
$\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \dfrac{6t}{2} = 3t$
$\dfrac{d^2y}{dx^2} = \dfrac{3t}{2}$
At $t=2$: $\dfrac{dy}{dx} = \dfrac{9}{2} \neq 0$, so $t=2$ is not a stationary point — it is simply a point on the curve where the curvature is $\dfrac{d^2y}{dx^2} = 3 > 0$ (concave up).