Derivatives

The Derivative as a Limit (CED Unit 2)

The derivative of $f$ at $x = a$ is defined as:

Adjust the parameters in the graph above to explore the relationships between variables.

F'(a) = \lim_\{h \to 0\} \frac\{f(a+h) - f(a)\}\{h\}

Equivalently, using an alternate form:

F'(a) = \lim_\{x \to a\} \frac\{f(x) - f(a)\}\{x - a\}

If this limit exists, $f$ is said to be differentiable at $a$ .

Physical intuition. If $s(t)$ is position at time $t$ Then $s'(t)$ is the instantaneous Velocity. The derivative answers: “how fast is $f$ changing right now?”

Geometric intuition. The derivative $f'(a)$ equals the slope of the tangent line to the graph of $f$ at the point $(a, f(a))$ .

Interpretations of the Derivative

Geometric: $f'(a)$ is the slope of the tangent line to $y = f(x)$ at the point $(a, f(a))$ .
Physical: If $s(t)$ is position, then $s'(t)$ is instantaneous velocity.
Rate of change: $f'(a)$ gives the instantaneous rate of change of $f$ with respect to $x$ at $x = a$ .

Find $f'(x)$ for $f(x) = \sqrt{x}$ using the limit definition.

F'(x) = \lim_\{h \to 0\} \frac\{\sqrt\{x+h\} - \sqrt\{x\}\}\{h\}

Multiply numerator and denominator by $\sqrt{x+h} + \sqrt{x}$ :

= \lim_\{h \to 0\} \frac\{x + h - x\}\{h(\sqrt\{x+h\} + \sqrt\{x\})\} = \lim_\{h \to 0\} \frac\{1\}\{\sqrt\{x+h\} + \sqrt\{x\}\} = \frac\{1\}\{2\sqrt\{x\}\}

This confirms the power rule for $n = \frac{1}{2}$ : $\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ . $\blacksquare$

Differentiability and Continuity

Theorem. If $f$ is differentiable at $a$ Then $f$ is continuous at $a$ .

Proof. We need to show $\displaystyle\lim_{x \to a} f(x) = f(a)$ . Consider:

\lim_\{x \to a\} [f(x) - f(a)] = \lim_\{x \to a\} \frac\{f(x) - f(a)\}\{x - a\} \cdot (x - a) = f'(a) \cdot 0 = 0

Therefore, $\displaystyle\lim_{x \to a} f(x) = f(a)$ So $f$ is continuous at $a$ . $\blacksquare$

The converse is false: $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there Because the left-hand and right-hand derivatives disagree.

Where Differentiability Fails

A function fails to be differentiable at a point if:

There is a corner (left and right derivatives differ), e.g., $|x|$ at $x = 0$ .
There is a cusp, e.g., $f(x) = x^{2/3}$ at $x = 0$ .
There is a vertical tangent, e.g., $f(x) = \sqrt[3]{x}$ at $x = 0$ .
There is a discontinuity.

Proof that $|x|$ is not differentiable at $x = 0$ .

Left-hand derivative: $\displaystyle\lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$ .

Right-hand derivative: $\displaystyle\lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$ .

Since $-1 \neq 1$ The derivative does not exist. $\blacksquare$

Proof that $f(x) = x^{2/3}$ has a cusp at $x = 0$ .

F'(x) = \frac\{2\}\{3\}x^\{-1/3\} = \frac\{2\}\{3\sqrt[3]\{x\}\}

As $x \to 0^+$$f'(x) \to +\infty$ . As $x \to 0^-$$f'(x) \to -\infty$ .

The left and right derivatives are both infinite but with opposite signs, creating a cusp. The Tangent line approaches vertical from both sides. $\blacksquare$

Differentiation Rules

Basic Rules

Let $c$ be a constant and $n$ a real number:

Constant rule: $\displaystyle\frac{d}{dx}[c] = 0$
Power rule: $\displaystyle\frac{d}{dx}[x^n] = nx^{n-1}$
Constant multiple rule: $\displaystyle\frac{d}{dx}[cf(x)] = cf'(x)$
Sum/Difference rule: $\displaystyle\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)$

Product Rule

\frac\{d\}\{dx\}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)

Proof sketch. Write $F(x) = f(x)g(x)$ and expand the difference quotient $\frac{F(x+h) - F(x)}{h}$ by adding and subtracting $f(x+h)g(x)$ :

\frac\{f(x+h)g(x+h) - f(x)g(x)\}\{h\} = f(x+h)\frac\{g(x+h) - g(x)\}\{h\} + g(x)\frac\{f(x+h) - f(x)\}\{h\}

Taking limits as $h \to 0$ gives the result. $\blacksquare$

Quotient Rule

\frac\{d\}\{dx\}\!\left[\frac\{f(x)\}\{g(x)\}\right] = \frac\{g(x)f'(x) - f(x)g'(x)\}\{[g(x)]^2\}

Proof. Let $F(x) = \frac{f(x)}{g(x)}$ So $f(x) = F(x)g(x)$ . Differentiating using the product Rule:

F'(x) = F'(x)g(x) + F(x)g'(x)

F'(x)g(x) = f'(x) - F(x)g'(x) = f'(x) - \frac\{f(x)\}\{g(x)\}g'(x)

F'(x) = \frac\{f'(x)\}\{g(x)\} - \frac\{f(x)g'(x)\}\{[g(x)]^2\} = \frac\{f'(x)g(x) - f(x)g'(x)\}\{[g(x)]^2\}

$\blacksquare$

Mnemonic: “Low d-High minus High d-Low, draw a line and square below.”

Chain Rule

\frac\{d\}\{dx\}[f(g(x))] = f'(g(x)) \cdot g'(x)

Why the chain rule works. If $y = f(u)$ and $u = g(x)$ Then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ . This is a cancellation of the $du$ terms (formalised by limits).

Proof of the chain rule. Let $u = g(x)$ and define:

K(h) = \frac\{f(u + h) - f(u)\}\{h\} - f'(u) \quad \mathrm\{when h \ne 0, \quad k(0) = 0

Then $\displaystyle\lim_{h \to 0} k(h) = 0$ and $f(u+h) - f(u) = [f'(u) + k(h)] \cdot h$ for all $h$ .

Let $\Delta u = g(x+h) - g(x)$ . Then:

\frac\{f(g(x+h)) - f(g(x))\}\{h\} = \frac\{f(u + \Delta u) - f(u)\}\{h\} = [f'(u) + k(\Delta u)] \cdot \frac\{\Delta u\}\{h\}

Taking $h \to 0$ : $\Delta u \to 0$ (since $g$ is differentiable, hence continuous), so $k(\Delta u) \to 0$ Giving:

F'(g(x)) \cdot g'(x)

$\blacksquare$

Find $\displaystyle\frac{d}{dx}\left[e^{\sqrt{3x+1}}\right]$ .

Apply the chain rule twice:

\frac\{d\}\{dx\}\left[e^\{\sqrt\{3x+1\}\}\right] = e^\{\sqrt\{3x+1\}\} \cdot \frac\{d\}\{dx\}\!\left[(3x+1)^\{1/2\}\right] = e^\{\sqrt\{3x+1\}\} \cdot \frac\{1\}\{2\}(3x+1)^\{-1/2\} \cdot 3 = \frac\{3e^\{\sqrt\{3x+1\}\}\}\{2\sqrt\{3x+1\}\}

Derivatives of Trigonometric Functions

\frac\{d\}\{dx\}[\sin x] = \cos x \qquad \frac\{d\}\{dx\}[\cos x] = -\sin x

\frac\{d\}\{dx\}[\tan x] = \sec^2 x \qquad \frac\{d\}\{dx\}[\cot x] = -\csc^2 x

\frac\{d\}\{dx\}[\sec x] = \sec x \tan x \qquad \frac\{d\}\{dx\}[\csc x] = -\csc x \cot x

Proof that $\displaystyle\frac{d}{dx}[\sin x] = \cos x$

\frac\{d\}\{dx\}[\sin x] = \lim_\{h \to 0\} \frac\{\sin(x+h) - \sin x\}\{h\} = \lim_\{h \to 0\} \frac\{\sin x \cos h + \cos x \sin h - \sin x\}\{h\}

= \lim_\{h \to 0\}\!\left[\sin x \cdot \frac\{\cos h - 1\}\{h\} + \cos x \cdot \frac\{\sin h\}\{h\}\right] = \sin x \cdot 0 + \cos x \cdot 1 = \cos x

Proof that $\displaystyle\frac{d}{dx}[\tan x] = \sec^2 x$

\frac\{d\}\{dx\}[\tan x] = \frac\{d\}\{dx\}\!\left[\frac\{\sin x\}\{\cos x\}\right] = \frac\{\cos x \cdot \cos x - \sin x \cdot (-\sin x)\}\{\cos^2 x\} = \frac\{\cos^2 x + \sin^2 x\}\{\cos^2 x\} = \frac\{1\}\{\cos^2 x\} = \sec^2 x

$\blacksquare$

Derivatives of Exponential and Logarithmic Functions

\frac\{d\}\{dx\}[e^x] = e^x \qquad \frac\{d\}\{dx\}[a^x] = a^x \ln a

\frac\{d\}\{dx\}[\ln x] = \frac\{1\}\{x\} \qquad \frac\{d\}\{dx\}[\log_a x] = \frac\{1\}\{x \ln a\}

Proof that $\displaystyle\frac{d}{dx}[e^x] = e^x$

\frac\{d\}\{dx\}[e^x] = \lim_\{h \to 0\} \frac\{e^\{x+h\} - e^x\}\{h\} = e^x \lim_\{h \to 0\} \frac\{e^h - 1\}\{h\} = e^x \cdot 1 = e^x

Where we used $\displaystyle\lim_{h \to 0} \frac{e^h - 1}{h} = 1$ .

Proof that $\displaystyle\frac{d}{dx}[\ln x] = \frac{1}{x}$

Let $y = \ln x$ So $e^y = x$ . Differentiating implicitly: $e^y \frac{dy}{dx} = 1$ So $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$ . $\blacksquare$

Proof that $\displaystyle\frac{d}{dx}[a^x] = a^x \ln a$

Write $a^x = e^{x \ln a}$ . Then:

\frac\{d\}\{dx\}[a^x] = \frac\{d\}\{dx\}[e^\{x \ln a\}] = e^\{x \ln a\} \cdot \ln a = a^x \ln a

$\blacksquare$

Inverse Trigonometric Derivatives

\frac\{d\}\{dx\}[\arcsin x] = \frac\{1\}\{\sqrt\{1 - x^2\}\} \qquad \frac\{d\}\{dx\}[\arccos x] = \frac\{-1\}\{\sqrt\{1 - x^2\}\}

\frac\{d\}\{dx\}[\arctan x] = \frac\{1\}\{1 + x^2\}

Derivation of $\arcsin x$ . Let $y = \arcsin x$ So $x = \sin y$ . Differentiating: $1 = \cos y \cdot \frac\{dy\}\{dx\}$ , so $\frac\{dy\}\{dx\} = \frac\{1\}\{\cos y\} = \frac\{1\}\{\sqrt\{1 - \sin^2 y\}\} = \frac\{1\}\{\sqrt\{1 - x^2\}\}$ .

Derivation of $\arctan x$ . Let $y = \arctan x$ So $x = \tan y$ . Differentiating: $1 = \sec^2 y \cdot \frac{dy}{dx}$ So $\frac{dy}{dx} = \frac{1}{\sec^2 y} = \cos^2 y$ .

Since $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ We get $\cos^2 y = \frac{1}{1 + x^2}$ So $\frac{dy}{dx} = \frac{1}{1 + x^2}$ . $\blacksquare$

Implicit Differentiation

When $y$ is defined implicitly as a function of $x$ Differentiate both sides with respect to $x$ And solve for $\frac{dy}{dx}$ .

Find $\displaystyle\frac{dy}{dx}$ for $e^{xy} + y = x^2$ .

Differentiate implicitly with respect to $x$ :

E^\{xy\}\!\left(y + x\frac\{dy\}\{dx\}\right) + \frac\{dy\}\{dx\} = 2x

E^\{xy\} \cdot y + e^\{xy\} \cdot x \frac\{dy\}\{dx\} + \frac\{dy\}\{dx\} = 2x

\frac\{dy\}\{dx\}(xe^\{xy\} + 1) = 2x - ye^\{xy\}

\frac\{dy\}\{dx\} = \frac\{2x - ye^\{xy\}\}\{xe^\{xy\} + 1\}

Higher-Order Derivatives

The second derivative is the derivative of the first derivative:

F''(x) = \frac\{d\}\{dx\}[f'(x)] = \frac\{d^2y\}\{dx^2\}

Similarly, $f'''(x)$$f^{(4)}(x)$ Etc. Denote higher-order derivatives.

Interpretation of the Second Derivative

If $f''(x) \gt 0$$f$ is concave up at $x$ .
If $f''(x) \lt 0$$f$ is concave down at $x$ .
If $f''(x)$ changes sign at $x = c$ Then $(c, f(c))$ is an inflection point.

Theorem. If $f$ has a local maximum at $c$ and $f''(c)$ exists, then $f''(c) \leq 0$ . If $f$ has A local minimum at $c$ and $f''(c)$ exists, then $f''(c) \geq 0$ .

This is the second derivative test for concavity at critical points.

Notation

Common notations for derivatives include:

$f'(x)$$f''(x)$$f'''(x)$ (prime notation)
$\frac{df}{dx}$$\frac{d^2f}{dx^2}$ (Leibniz notation)
$\dot{x}$$\ddot{x}$ (Newton’s dot notation for time derivatives)

Applications of Derivatives (CED Unit 5)

When two or more quantities are related by an equation, their rates of change are also related.

A ladder 10 ft long rests against a vertical wall. If the bottom slides away at 1 ft/s, how fast is the top sliding down when the bottom is 6 ft from the wall?

Let $x$ = distance from wall to bottom, $y$ = height of top on wall. Then $x^2 + y^2 = 100$ .

Differentiate with respect to $t$ :

2x\frac\{dx\}\{dt\} + 2y\frac\{dy\}\{dt\} = 0

When $x = 6$ : $y = \sqrt{100 - 36} = 8$ . Given $\frac{dx}{dt} = 1$ :

2(6)(1) + 2(8)\frac\{dy\}\{dt\} = 0 \implies \frac\{dy\}\{dt\} = -\frac\{12\}\{16\} = -\frac\{3\}\{4\} \mathrm\{ ft/s

The negative sign confirms the top is sliding down.

A 15 ft ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the angle between the ladder and the ground changing when the bottom is 9 ft from the wall?

Let $\theta$ be the angle between the ladder and the ground. Then $\cos\theta = \frac{x}{15}$ So $x = 15\cos\theta$ .

Differentiating with respect to $t$ :

\frac\{dx\}\{dt\} = -15\sin\theta \frac\{d\theta\}\{dt\}

When $x = 9$ : $\cos\theta = \frac{9}{15} = \frac{3}{5}$ So $\sin\theta = \frac{4}{5}$ .

2 = -15 \cdot \frac\{4\}\{5\} \cdot \frac\{d\theta\}\{dt\} = -12 \frac\{d\theta\}\{dt\}

\frac\{d\theta\}\{dt\} = -\frac\{2\}\{12\} = -\frac\{1\}\{6\} \mathrm\{ rad/s

The angle is decreasing at $\frac{1}{6}$ rad/s.

Linear Approximation and Differentials

The tangent line at $x = a$ gives a linear approximation:

F(x) \approx f(a) + f'(a)(x - a)

This is useful for estimating values of functions near known points.

Use linear approximation to estimate $(1.02)^{10}$ .

Let $f(x) = x^{10}$$a = 1$ .

F'(x) = 10x^9 \implies f(1) = 1, \quad f'(1) = 10

(1.02)^\{10\} \approx 1 + 10(0.02) = 1.20

(The actual value is approximately 1.21899, showing that linear approximation is less accurate when $x$ is farther from $a$ .)

Mean Value Theorem (MVT)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ Then there exists at least one $c \in (a, b)$ such that:

F'(c) = \frac\{f(b) - f(a)\}\{b - a\}

This means the instantaneous rate of change equals the average rate of change at some interior Point.

Geometric interpretation. The MVT guarantees the existence of a tangent line parallel to the Secant line joining $(a, f(a))$ and $(b, f(b))$ .

Proof of the MVT. Let $g(x) = f(x) - \frac{f(b) - f(a)}{b-a}(x - a) - f(a)$ . Then $g(a) = g(b) = 0$ . By Rolle’s Theorem, there exists $c \in (a, b)$ with $g'(c) = 0$ Which gives $f'(c) = \frac{f(b) - f(a)}{b-a}$ . $\blacksquare$

Rolle’s Theorem

If $f$ is continuous on $[a, b]$ Differentiable on $(a, b)$ And $f(a) = f(b)$ Then there exists At least one $c \in (a, b)$ such that $f'(c) = 0$ .

Rolle’s Theorem is a special case of the MVT where $f(a) = f(b)$ .

Application of Rolle’s Theorem. Rolle’s Theorem is used to prove that a differentiable function Has at most one root in an interval. If it had two roots, Rolle’s Theorem would give a point where the derivative is zero, leading to a contradiction.

Show that $f(x) = x^3 - 3x + 1$ has exactly one real root.

Note that $f(0) = 1$ and $f(-2) = -8 + 6 + 1 = -1$ . By the IVT, there is at least one root in $(-2, 0)$ .

Suppose there are two distinct roots $a \lt b$ . Then $f(a) = f(b) = 0$ And by Rolle’s Theorem, There exists $c \in (a, b)$ with $f'(c) = 0$ . But $f'(x) = 3x^2 - 3 = 3(x+1)(x-1)$ Which is zero Only at $x = \pm 1$ . This does not lead to a contradiction on its own, so we must examine further.

In fact, $f(-2) = -1$$f(0) = 1$$f(1) = -1$$f(2) = 3$ . So $f$ has at least one root in $(-2, 0)$ At least one in $(0, 1)$ And possibly one in $(1, 2)$ . Rolle’s Theorem applied to Consecutive roots shows that each pair of roots must bracket either $x = -1$ or $x = 1$ .

Testing: $f(-1) = -1 + 3 + 1 = 3 \gt 0$ . So the root in $(-2, 0)$ is in $(-2, -1)$ And the root in $(0, 1)$ is in $(0, 1)$ . These bracket different critical points. In fact, $f$ has three real roots, And Rolle’s Theorem confirms this is consistent since the critical points at $x = \pm 1$ bracket the Three roots.

Curve Sketching and Analysis

Critical Points

A critical number of $f$ is a value $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does not exist.

First Derivative Test

Let $c$ be a critical number:

If $f'$ changes from positive to negative at $c$ Then $f(c)$ is a local maximum.
If $f'$ changes from negative to positive at $c$ Then $f(c)$ is a local minimum.
If $f'$ does not change sign at $c$ Then $f(c)$ is neither.

Second Derivative Test

Let $f''(c)$ exist with $f'(c) = 0$ :

If $f''(c) \gt 0$ Then $f(c)$ is a local minimum.
If $f''(c) \lt 0$ Then $f(c)$ is a local maximum.
If $f''(c) = 0$ The test is inconclusive.

When to use which test. The first derivative test always works. The second derivative test is Faster but sometimes inconclusive. When $f''(c) = 0$ You must fall back to the first derivative Test.

Example where the second derivative test is inconclusive. Consider $f(x) = x^4$ . $f'(x) = 4x^3 = 0$ gives $x = 0$ And $f''(x) = 12x^2$ So $f''(0) = 0$ — the test is inconclusive. But $f'(x) = 4x^3$ changes from negative to positive at $x = 0$ So $f$ has a local minimum at $x = 0$ by the first derivative test.

Optimization

To solve optimization problems:

Identify the quantity to optimize and the constraint.
Write an equation relating the variables.
Express the quantity as a function of a single variable.
Find critical points and apply the first or second derivative test.
Verify the result answers the original question.

A rectangular box with a square base has volume $500 \mathrm{ cm^3}$ . The material for the base costs Twice as much as the material for the sides. Minimise the cost of the material.

Let the base have side length $x$ cm and height $h$ cm. Let the cost per unit area of the sides be $c$ . Then the base costs $2c$ per unit area.

Volume: $x^2 h = 500$ So $h = \frac{500}{x^2}$ .

Cost: $C = 2c \cdot x^2 + 4c \cdot xh = 2cx^2 + 4cx \cdot \frac{500}{x^2} = 2cx^2 + \frac{2000c}{x}$ .

\frac\{dC\}\{dx\} = 4cx - \frac\{2000c\}\{x^2\} = 0

4x^3 = 2000 \implies x^3 = 500 \implies x = \sqrt[3]\{500\} \approx 7.937 \mathrm\{ cm

H = \frac\{500\}\{500^\{2/3\}\} = 500^\{1/3\} = \sqrt[3]\{500\} \approx 7.937 \mathrm\{ cm

The cost is minimised when the box is a cube. $\frac{d^2C}{dx^2} = 4c + \frac{4000c}{x^3} \gt 0$ Confirming a minimum.

Optimization on Closed Intervals

When the domain is a closed interval $[a, b]$ The absolute maximum and minimum occur at critical Points or at endpoints.

Procedure:

Find all critical points in $(a, b)$ .
Evaluate $f$ at all critical points and at both endpoints.
The largest value is the absolute max; the smallest is the absolute min.

Common Pitfalls

Forgetting the chain rule. When differentiating a composition, always account for the inner function’s derivative. For example, $\frac{d}{dx}[\sin(x^2)] \ne \cos(x^2)$ .
Incorrect sign in the quotient rule. Remember: “Low d-High minus High d-Low, draw a line and square below.”
Confusing $\frac{d}{dx}[e^x]$ with $\frac{d}{dx}[x^e]$ . The former is $e^x$ ; the latter is $ex^{e-1}$ .
Implicit differentiation: forgetting to apply the chain rule to $y$ terms. When differentiating $y^2$ with respect to $x$ The result is $2y\frac{dy}{dx}$ Not $2y$ .
Units in related rates. Always include units and check that they make dimensional sense.
Not checking endpoints in optimization problems on closed intervals.
Using the second derivative test when it is inconclusive ( $f''(c) = 0$ ). Fall back to the first derivative test.
Confusing the second derivative with the first. The second derivative tells you about concavity, not the slope.
Dropping negative signs in implicit differentiation. Be systematic: write every $y$ term with $\frac{dy}{dx}$ attached.
Linear approximation overreach. The approximation $f(x) \approx f(a) + f'(a)(x-a)$ is accurate only near $x = a$ . Using it far from $a$ can give wildly inaccurate results.

Practice Questions

Find $\displaystyle\frac{dy}{dx}$ for $e^{xy} + y = x^2$ using implicit differentiation.
A spherical balloon is inflated at a rate of $10 \mathrm{ cm^3/\mathrm{s}}$ . How fast is the radius increasing when the radius is 5 cm? ( $V = \frac{4}{3}\pi r^3$ )
Find all critical points of $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$ and classify them using the second derivative test.
Use the Mean Value Theorem to show that $|\sin b - \sin a| \le |b - a|$ for all $a, b$ .
Find the equation of the tangent line to $y = \ln(x^2 + 1)$ at $x = 1$ .
A 15 ft ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the angle between the ladder and the ground changing when the bottom is 9 ft from the wall?
Find the absolute maximum and minimum of $f(x) = x^3 - 3x^2 + 4$ on $[-1, 4]$ .
Use linear approximation to estimate $(1.02)^{10}$ .
Find $\displaystyle\frac{d}{dx}\left[\frac{x^2 + 1}{e^{2x}}\right]$ and find all stationary points.
Find $\displaystyle\frac{d^2y}{dx^2}$ for $x^3 + y^3 = 6xy$ at the point $(3, 3)$ .
A rectangular box with a square base has volume $500 \mathrm{ cm^3}$ . The material for the base costs twice as much as the material for the sides. Minimise the cost of the material.
Prove that the derivative of $f(x) = x^3$ is $f'(x) = 3x^2$ from first principles.
Find $\displaystyle\frac{dy}{dx}$ for $\sin(x + y) = x \cos y$ .
Find the inflection points of $f(x) = x^4 - 6x^2 + 4$ .
A conical tank (vertex pointing down) has radius 5 ft at the top and height 10 ft. Water is pumped in at 3 $\mathrm{ft^3/\mathrm{min}}$ . How fast is the water level rising when the water is 6 ft deep?
Find the dimensions of the rectangle of maximum area that can be inscribed in a semicircle of radius $r$ .

Practice Problems

Question 1: Implicit differentiation

Find $\frac{dy}{dx}$ for the curve $x^3 + y^3 = 6xy$ . Then find the equation of the tangent line at the point $(3, 3)$ .

Answer

Differentiate implicitly: $3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$ .

$3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$ .

$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$ .

At $(3, 3)$ : $\frac{dy}{dx} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1$ .

Tangent line: $y - 3 = -1(x - 3)$ So $y = -x + 6$ .

Question 2: Related rates

A spherical balloon is being inflated at a rate of $100 \mathrm{ cm^3/s}$ . How fast is the radius increasing when the radius is $10 \mathrm{ cm}$ ?

Answer

$V = \frac{4}{3}\pi r^3$ . Differentiate with respect to time: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ .

$100 = 4\pi(10)^2 \frac{dr}{dt} = 400\pi \frac{dr}{dt}$ .

$\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.080 \mathrm{ cm/s}$ .

Question 3: L’Hopital’s rule

Evaluate $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ .

Answer

This is $0/0$ . Apply L’Hopital’s rule:

$\displaystyle\lim_{x \to 0} \frac{e^x - 1}{2x}$ .

Still $0/0$ . Apply again:

$\displaystyle\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$ .

Alternatively, this is the coefficient of $x^2/2!$ in the Taylor series of $e^x$ Confirming the answer.

Question 4: Optimisation with constraint

A rectangular box with a square base and no top is to have a volume of $108 \mathrm{ cm^3}$ . Find the dimensions that minimise the surface area.

Answer

Let base side = $x$ Height = $h$ . Volume: $x^2 h = 108$ So $h = 108/x^2$ .

Surface area: $S = x^2 + 4xh = x^2 + 4x(108/x^2) = x^2 + 432/x$ .

$\frac{dS}{dx} = 2x - 432/x^2 = 0$ So $2x = 432/x^2$$x^3 = 216$$x = 6 \mathrm{ cm}$ .

$h = 108/36 = 3 \mathrm{ cm}$ .

$\frac{d^2S}{dx^2} = 2 + 864/x^3 \gt 0$ at $x = 6$ Confirming minimum.

Dimensions: $6 \times 6 \times 3 \mathrm{ cm}$ Minimum surface area $= 36 + 432/6 = 108 \mathrm{ cm^2}$ .

Question 5: Mean Value Theorem

Verify that the function $f(x) = x^3 - 3x + 1$ satisfies the conditions of the Mean Value Theorem on $[-2, 2]$ And find all values of $c$ that satisfy the conclusion.

Answer

$f$ is a polynomial, so it is continuous on $[-2, 2]$ and differentiable on $(-2, 2)$ .

$f(2) = 8 - 6 + 1 = 3$ . $f(-2) = -8 + 6 + 1 = -1$ .

$f'(c) = \frac{f(2) - f(-2)}{2 - (-2)} = \frac{3 - (-1)}{4} = 1$ .

$f'(x) = 3x^2 - 3 = 1$ So $3x^2 = 4$$x^2 = 4/3$$x = \pm 2/\sqrt{3} \approx \pm 1.155$ .

Both values are in $(-2, 2)$ . The MVT is satisfied at $c = 2/\sqrt{3}$ and $c = -2/\sqrt{3}$ .

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Summary

This topic covers the mathematical techniques and concepts related to derivatives, including key theorems, methods, and problem-solving approaches.

Key concepts include:

sine, cosine, and tangent functions
trigonometric identities
solving trigonometric equations
the sine and cosine rules
radian measure and arc length

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Intuition

The speedometer analogy: A derivative tells you how fast something is changing at any instant, just like a speedometer shows your current speed without needing to measure average speed over a distance.

Why it matters: Derivatives are everywhere, from finding the optimal price for a product to understanding how diseases spread. They allow us to model and predict change in any system.

The key insight: A derivative captures the instantaneous rate of change by shrinking the measurement interval to zero.

Cross-References

Limits and Continuity — The derivative is defined as a limit, making limits the foundational concept for all differentiation.
Integrals — Integration is the reverse process of differentiation, connecting the two fundamental operations of calculus.
Differential Equations — Derivatives appear in differential equations, linking rate-of-change analysis to modelling real-world phenomena.
AP Physics, Kinematics: Velocity is the derivative of position, and acceleration is the derivative of velocity, derivatives model all rates of change in mechanics.
AP Physics, Newton’s Laws: Force equals mass times acceleration, where acceleration is a second derivative of position, connecting differentiation directly to dynamics.

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