4.4 Application Of Derivatives

Detailed Theory: Applications of Derivatives

1. Introduction to Applications of Derivatives

1.1 Why Study Applications?

Derivatives are not just abstract mathematical concepts; they have powerful real-world applications in:

• Finding maximum/minimum values (optimization)

• Analyzing rates of change

• Understanding curve behavior

• Solving physics and engineering problems

1.2 The Derivative as a Tool

The derivative $f'(x)$ gives us:

• Slope of the tangent line at point $x$

• Rate of change of $f(x)$ with respect to $x$

• Instantaneous velocity if $f(t)$ represents position

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2. Tangent and Normal Lines

2.1 Tangent Line

The line that just "touches" a curve at a point, having the same slope as the curve at that point.

Equation at point $(x_0, y_0)$: $y - y_0 = f'(x_0)(x - x_0)$

2.2 Normal Line

The line perpendicular to the tangent line at the point of tangency.

Equation at point $(x_0, y_0)$: $y - y_0 = -\frac{1}{f'(x_0)}(x - x_0)$ (if $f'(x_0) \neq 0$)

Special case: If $f'(x_0) = 0$, tangent is horizontal, normal is vertical: $x = x_0$

2.3 Example

Find tangent and normal to $y = x^2$ at $(1, 1)$

$$f'(x) = 2x \Rightarrow f'(1) = 2$$

Tangent line: $y - 1 = 2(x - 1) \Rightarrow y = 2x - 1$

Normal line: $y - 1 = -\frac{1}{2}(x - 1) \Rightarrow y = -\frac{1}{2}x + \frac{3}{2}$

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3. Increasing and Decreasing Functions

3.1 Definitions

• Increasing: $f(x_1) < f(x_2)$ whenever $x_1 < x_2$

• Decreasing: $f(x_1) > f(x_2)$ whenever $x_1 < x_2$

• Strictly increasing/decreasing: No equalities allowed

3.2 Test Using Derivatives

• If $f'(x) > 0$ on interval $I$, then $f$ is increasing on $I$

• If $f'(x) < 0$ on interval $I$, then $f$ is decreasing on $I$

• If $f'(x) = 0$ on interval $I$, then $f$ is constant on $I$

3.3 Finding Intervals of Increase/Decrease

Steps:

1. Find critical points where $f'(x) = 0$ or $f'(x)$ undefined

2. Use number line to test sign of $f'(x)$ in each interval

3.4 Example

Find intervals where $f(x) = x^3 - 3x^2 - 9x + 5$ is increasing/decreasing

$$f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$$

Critical points: $x = -1$, $x = 3$

Test intervals:

• $(-\infty, -1)$: Test $x = -2$: $f'(-2) = 3(-5)(-3) = 45 > 0$ (increasing)

• $(-1, 3)$: Test $x = 0$: $f'(0) = 3(-3)(1) = -9 < 0$ (decreasing)

• $(3, \infty)$: Test $x = 4$: $f'(4) = 3(1)(5) = 15 > 0$ (increasing)

So: Increasing on $(-\infty, -1) \cup (3, \infty)$, decreasing on $(-1, 3)$

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4. Local Maxima and Minima (First Derivative Test)

4.1 Critical Points

A point $c$ in domain of $f$ is a critical point if:

1. $f'(c) = 0$, or

2. $f'(c)$ does not exist

Important: All local maxima/minima occur at critical points, but not all critical points give maxima/minima.

4.2 First Derivative Test

For critical point $c$:

1. If $f'$ changes from positive to negative at $c$: Local maximum at $c$

2. If $f'$ changes from negative to positive at $c$: Local minimum at $c$

3. If $f'$ does not change sign at $c$: Neither maximum nor minimum (inflection point)

4.3 Example

Find local extrema of $f(x) = x^3 - 3x^2 - 9x + 5$

From previous: $f'(x) = 3(x-3)(x+1)$

Critical points: $x = -1$, $x = 3$

Sign analysis:

• Left of $-1$: $f' > 0$ (increasing)

• Right of $-1$: $f' < 0$ (decreasing)

• So $x = -1$ is local maximum

• Left of $3$: $f' < 0$ (decreasing)

• Right of $3$: $f' > 0$ (increasing)

• So $x = 3$ is local minimum

Values: $f(-1) = 10$ (local max), $f(3) = -22$ (local min)

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5. Concavity and Inflection Points (Second Derivative Test)

5.1 Concavity

• Concave up: Graph lies above tangent lines (shaped like ∪)

• Concave down: Graph lies below tangent lines (shaped like ∩)

5.2 Test Using Second Derivative

• If $f''(x) > 0$ on interval $I$: $f$ is concave up on $I$

• If $f''(x) < 0$ on interval $I$: $f$ is concave down on $I$

5.3 Inflection Points

A point where concavity changes (from up to down or down to up).

To find inflection points:

1. Find where $f''(x) = 0$ or $f''(x)$ undefined

2. Check if concavity changes at those points

5.4 Second Derivative Test for Extrema

For critical point $c$ with $f'(c) = 0$:

1. If $f''(c) > 0$: Local minimum at $c$

2. If $f''(c) < 0$: Local maximum at $c$

3. If $f''(c) = 0$: Test inconclusive (use First Derivative Test)

5.5 Example

Analyze $f(x) = x^4 - 4x^3$

First derivative: $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$

Critical points: $x = 0$, $x = 3$

Second derivative: $f''(x) = 12x^2 - 24x = 12x(x - 2)$

At $x = 3$: $f''(3) = 12(3)(1) = 36 > 0$ ⇒ Local minimum

At $x = 0$: $f''(0) = 0$ ⇒ Test inconclusive

Use First Derivative Test at $x = 0$:

• Both sides have $f' < 0$ (decreasing)

• So $x = 0$ is not extremum

Concavity: $f''(x) = 0$ at $x = 0$, $x = 2$

Test intervals:

• $(-\infty, 0)$: Test $x = -1$, $f''(-1) = 12(-1)(-3) = 36 > 0$ (concave up)

• $(0, 2)$: Test $x = 1$, $f''(1) = 12(1)(-1) = -12 < 0$ (concave down)

• $(2, \infty)$: Test $x = 3$, $f''(3) = 12(3)(1) = 36 > 0$ (concave up)

Inflection points: $x = 0$ and $x = 2$

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6. Curve Sketching

6.1 Steps for Curve Sketching

1. Domain: Find where function is defined

2. Intercepts: Find x-intercepts ($f(x)=0$) and y-intercept ($f(0)$)

3. Symmetry: Check if even ($f(-x)=f(x)$), odd ($f(-x)=-f(x)$), or periodic

4. Asymptotes: Vertical, horizontal, slant

5. Intervals of increase/decrease: Use first derivative

6. Local extrema: Identify maxima and minima

7. Concavity and inflection points: Use second derivative

8. Sketch: Plot key points and connect with appropriate shape

6.2 Example: Sketch $f(x) = \frac{x^2}{x^2-1}$

1. Domain: All real $x$ except $x = \pm 1$

2. Intercepts: $f(0)=0$ (y-intercept), $x=0$ (x-intercept)

3. Symmetry: $f(-x) = f(x)$ ⇒ even function (symmetric about y-axis)

4. Asymptotes:

   • Vertical: $x = 1$ and $x = -1$ (denominator = 0)

   • Horizontal: As $x \to \pm\infty$, $f(x) \to 1$ ⇒ $y=1$

5. First derivative: $f'(x) = \frac{2x(x^2-1) - x^2(2x)}{(x^2-1)^2} = \frac{-2x}{(x^2-1)^2}$

   Critical point: $x=0$ (where $f'(x)=0$)

   Sign: $f'(x) > 0$ for $x<0$, $f'(x) < 0$ for $x>0$

6. Increasing: $(-\infty, -1) \cup (-1, 0)$ Decreasing: $(0, 1) \cup (1, \infty)$

7. Local extremum: Local maximum at $x=0$ ($f(0)=0$)

8. Second derivative: $f''(x) = \frac{2(3x^2+1)}{(x^2-1)^3}$

   Always positive except at discontinuities ⇒ Always concave up

9. Sketch: Symmetric bell shape with vertical asymptotes at ±1, horizontal asymptote at y=1

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7. Optimization Problems

7.1 Strategy for Optimization

1. Understand: Read problem carefully, identify quantity to optimize

2. Draw diagram: If applicable, sketch the situation

3. Formulate: Write equation for quantity to optimize (objective function)

4. Constraint: Write constraint equation(s)

5. Reduce variables: Use constraint to express objective in one variable

6. Domain: Determine practical domain

7. Find critical points: Take derivative, set to zero

8. Check endpoints: Evaluate at critical points and endpoints

9. Interpret: Answer in context of problem

7.2 Example 1: Maximum Area Problem

Find rectangle of maximum area that can be inscribed in semicircle of radius $r$

Solution: Let rectangle have width $2x$ and height $y$

Constraint: $x^2 + y^2 = r^2$ (from semicircle equation)

Objective: Maximize area $A = 2xy$

Express in one variable: $y = \sqrt{r^2 - x^2}$

$A(x) = 2x\sqrt{r^2 - x^2}$, domain: $0 \leq x \leq r$

Find critical points:

$$A'(x) = 2\sqrt{r^2 - x^2} + 2x \cdot \frac{-x}{\sqrt{r^2 - x^2}} = \frac{2(r^2 - x^2) - 2x^2}{\sqrt{r^2 - x^2}}$$

$$= \frac{2r^2 - 4x^2}{\sqrt{r^2 - x^2}}$$

Set $A'(x) = 0$: $2r^2 - 4x^2 = 0 \Rightarrow x^2 = \frac{r^2}{2} \Rightarrow x = \frac{r}{\sqrt{2}}$

Check endpoints: $A(0) = 0$, $A(r) = 0$

Maximum at $x = \frac{r}{\sqrt{2}}$: $A_{\text{max}} = 2\left(\frac{r}{\sqrt{2}}\right)\sqrt{r^2 - \frac{r^2}{2}} = r^2$

7.3 Example 2: Minimize Surface Area

Find dimensions of cylindrical can (closed top and bottom) with volume $V$ that minimizes surface area

Solution: Let radius = $r$, height = $h$

Volume constraint: $V = \pi r^2 h \Rightarrow h = \frac{V}{\pi r^2}$

Surface area: $S = 2\pi r^2 + 2\pi rh$

Substitute $h$: $S(r) = 2\pi r^2 + 2\pi r\left(\frac{V}{\pi r^2}\right) = 2\pi r^2 + \frac{2V}{r}$

Domain: $r > 0$

Find critical points:

$$S'(r) = 4\pi r - \frac{2V}{r^2} = 0$$

$$4\pi r = \frac{2V}{r^2} \Rightarrow 4\pi r^3 = 2V \Rightarrow r^3 = \frac{V}{2\pi} \Rightarrow r = \sqrt[3]{\frac{V}{2\pi}}$$

Then $h = \frac{V}{\pi r^2} = \frac{V}{\pi\left(\frac{V}{2\pi}\right)^{2/3}} = \frac{V^{1/3}}{(\pi)^{1/3}}\left(\frac{2\pi}{V}\right)^{2/3} = 2\sqrt[3]{\frac{V}{2\pi}} = 2r$

Optimal ratio: $h = 2r$ (height = diameter)

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8. Related Rates

8.1 Concept

Problems where two or more related quantities are changing with time, and we want to find how fast one is changing given how fast others are changing.

8.2 Strategy

1. Identify variables: What quantities are changing? Assign variables

2. Given rates: What rates are known? $\frac{d?}{dt} = ?$

3. Wanted rate: What rate do we need to find? $\frac{d?}{dt} = ?$

4. Equation: Write equation relating variables

5. Differentiate: Differentiate with respect to time $t$

6. Substitute: Plug in known values

7. Solve: Solve for wanted rate

8.3 Example 1: Ladder Problem

A 10 ft ladder leans against wall. Bottom slides away at 1 ft/s. How fast is top sliding down when bottom is 6 ft from wall?

Solution: Let $x$ = distance from wall, $y$ = height on wall

Given: $\frac{dx}{dt} = 1$ ft/s Find: $\frac{dy}{dt}$ when $x = 6$

Equation: $x^2 + y^2 = 100$

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

When $x = 6$: $y = \sqrt{100 - 36} = 8$

Substitute: $2(6)(1) + 2(8)\frac{dy}{dt} = 0$

$12 + 16\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{3}{4}$ ft/s

Negative means sliding down.

8.4 Example 2: Spherical Balloon

Air is pumped into spherical balloon at rate of 100 cm³/s. How fast is radius increasing when radius is 10 cm?

Solution: Given: $\frac{dV}{dt} = 100$ cm³/s Find: $\frac{dr}{dt}$ when $r = 10$

Volume: $V = \frac{4}{3}\pi r^3$

Differentiate: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

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

$\frac{dr}{dt} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796$ cm/s

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9. Mean Value Theorems

9.1 Rolle's Theorem

If $f$ is:

1. Continuous on $[a,b]$

2. Differentiable on $(a,b)$

3. $f(a) = f(b)$

Then there exists $c$ in $(a,b)$ such that $f'(c) = 0$

Geometric meaning: Between two points with same height, there's a horizontal tangent.

9.2 Mean Value Theorem (MVT)

If $f$ is:

1. Continuous on $[a,b]$

2. Differentiable on $(a,b)$

Then there exists $c$ in $(a,b)$ such that:

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

Geometric meaning: There's a point where tangent slope equals secant slope.

9.3 Example Using MVT

Verify MVT for $f(x) = x^3 - x$ on $[0, 2]$

Check conditions: Polynomial ⇒ continuous and differentiable everywhere

$$\frac{f(2) - f(0)}{2 - 0} = \frac{(8-2) - 0}{2} = \frac{6}{2} = 3$$

Find $c$ such that $f'(c) = 3$

$f'(x) = 3x^2 - 1$

Set $3x^2 - 1 = 3 \Rightarrow 3x^2 = 4 \Rightarrow x^2 = \frac{4}{3} \Rightarrow x = \pm\frac{2}{\sqrt{3}}$

In $(0,2)$: $c = \frac{2}{\sqrt{3}} \approx 1.155$

9.4 Consequences of MVT

1. If $f'(x) = 0$ for all $x$ in $(a,b)$, then $f$ is constant on $(a,b)$

2. If $f'(x) > 0$ for all $x$ in $(a,b)$, then $f$ is increasing on $(a,b)$

3. If $f'(x) < 0$ for all $x$ in $(a,b)$, then $f$ is decreasing on $(a,b)$

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10. Linear Approximation and Differentials

10.1 Linear Approximation

Approximate function near point $a$ using tangent line:

$$L(x) = f(a) + f'(a)(x - a)$$

Approximation: $f(x) \approx L(x)$ for $x$ near $a$

10.2 Example

Approximate $\sqrt{16.1}$ using linear approximation

Let $f(x) = \sqrt{x}$, $a = 16$

$f(a) = \sqrt{16} = 4$

$f'(x) = \frac{1}{2\sqrt{x}} \Rightarrow f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8} = 0.125$

Linear approximation: $L(x) = 4 + 0.125(x - 16)$

For $x = 16.1$: $\sqrt{16.1} \approx 4 + 0.125(0.1) = 4 + 0.0125 = 4.0125$

Actual: $\sqrt{16.1} \approx 4.01248$ (very close!)

10.3 Differentials

If $y = f(x)$, then differential $dy = f'(x) dx$

Interpretation: $dy$ approximates change in $y$ when $x$ changes by $dx$

Example: Use differentials to estimate change in volume of sphere when radius increases from 10 cm to 10.1 cm

Volume: $V = \frac{4}{3}\pi r^3$

$dV = 4\pi r^2 dr$

Here $r = 10$, $dr = 0.1$

$dV = 4\pi(10)^2(0.1) = 40\pi \approx 125.66$ cm³

Actual change: $\Delta V = \frac{4}{3}\pi(10.1^3 - 10^3) = \frac{4}{3}\pi(1030.301 - 1000) \approx 127.17$ cm³

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11. L'Hôpital's Rule

11.1 Statement

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ has form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

provided the limit on right exists or is $\pm\infty$

11.2 Examples

Example 1: $\lim_{x \to 0} \frac{\sin x}{x}$ ($\frac{0}{0}$ form)

Apply L'Hôpital's: $\lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1$

Example 2: $\lim_{x \to \infty} \frac{x^2}{e^x}$ ($\frac{\infty}{\infty}$ form)

Apply L'Hôpital's twice:

First: $\lim_{x \to \infty} \frac{2x}{e^x}$

Second: $\lim_{x \to \infty} \frac{2}{e^x} = 0$

Example 3: $\lim_{x \to 0^+} x \ln x$ ($0 \cdot \infty$ form)

Rewrite: $\lim_{x \to 0^+} \frac{\ln x}{1/x}$ (now $\frac{\infty}{\infty}$)

Apply L'Hôpital's: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$

11.3 Other Indeterminate Forms

Can convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$:

1. $0 \cdot \infty$: Write as $\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$

2. $\infty - \infty$: Combine into single fraction

3. $0^0$, $\infty^0$, $1^\infty$: Take natural log

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12. Newton's Method for Finding Roots

12.1 The Method

Iterative method to approximate roots of $f(x) = 0$

Formula:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

12.2 Steps

1. Choose initial approximation $x_0$

2. Compute $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$

3. Repeat until desired accuracy

12.3 Example

Find approximate root of $f(x) = x^2 - 2 = 0$ (i.e., $\sqrt{2}$)

$f'(x) = 2x$

Choose $x_0 = 1$

Iteration 1: $x_1 = 1 - \frac{1^2 - 2}{2(1)} = 1 - \frac{-1}{2} = 1.5$

Iteration 2: $x_2 = 1.5 - \frac{(1.5)^2 - 2}{2(1.5)} = 1.5 - \frac{2.25 - 2}{3} = 1.5 - \frac{0.25}{3} \approx 1.4167$

Iteration 3: $x_3 = 1.4167 - \frac{(1.4167)^2 - 2}{2(1.4167)} \approx 1.4142$

$\sqrt{2} \approx 1.41421356$, so we're already close!

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13. Applications in Physics and Economics

13.1 Physics Applications

Motion along a line:

• Position: $s(t)$

• Velocity: $v(t) = s'(t)$

• Acceleration: $a(t) = v'(t) = s''(t)$

Example: Particle moves with $s(t) = t^3 - 6t^2 + 9t$

• Velocity: $v(t) = 3t^2 - 12t + 9$

• Acceleration: $a(t) = 6t - 12$

When is particle at rest? $v(t) = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow t = 1, 3$

Related rates in physics:

• Pressure and volume: Boyle's Law $PV = k$

• Force and work: $W = F \cdot d$

13.2 Economics Applications

Marginal Concepts:

• Marginal cost: $C'(x)$ ≈ cost of producing one more unit

• Marginal revenue: $R'(x)$ ≈ revenue from selling one more unit

• Marginal profit: $P'(x) = R'(x) - C'(x)$

Elasticity of Demand:

$$E(p) = \frac{p}{q} \cdot \frac{dq}{dp}$$

Interpretation:

• $|E| > 1$: Elastic (demand sensitive to price)

• $|E| = 1$: Unit elastic

• $|E| < 1$: Inelastic (demand not sensitive to price)

Example: Demand $q = 100 - 2p$

$$\frac{dq}{dp} = -2$$

$$E(p) = \frac{p}{100 - 2p} \cdot (-2) = -\frac{2p}{100 - 2p}$$

At $p = 20$: $E(20) = -\frac{40}{60} = -\frac{2}{3} \Rightarrow |E| = \frac{2}{3} < 1$ (inelastic)

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14. Important Theorems and Formulas

14.1 Key Theorems

1. Rolle's Theorem: $f(a)=f(b) \Rightarrow \exists c: f'(c)=0$

2. Mean Value Theorem: $\exists c: f'(c) = \frac{f(b)-f(a)}{b-a}$

3. First Derivative Test: Sign change of $f'$ indicates local extrema

4. Second Derivative Test: $f''(c)$ sign indicates concavity and extremum type

14.2 Optimization Formulas

1. Rectangle area: $A = lw$, perimeter $P = 2l + 2w$

2. Cylinder volume: $V = \pi r^2 h$, surface area $S = 2\pi r^2 + 2\pi rh$

3. Sphere volume: $V = \frac{4}{3}\pi r^3$, surface area $S = 4\pi r^2$

4. Cone volume: $V = \frac{1}{3}\pi r^2 h$

14.3 Related Rates Common Formulas

1. Pythagorean: $x^2 + y^2 = z^2$

2. Similar triangles: $\frac{a}{b} = \frac{c}{d}$

3. Volume formulas (see above)

4. Trigonometric relations in right triangles

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15. Solved Examples

Example 1: Complete Curve Analysis

Analyze $f(x) = \frac{x}{x^2+1}$

1. Domain: All real numbers (denominator never 0)

2. Intercepts: $f(0)=0$, only intercept

3. Symmetry: $f(-x) = -f(x)$ ⇒ odd function (symmetric about origin)

4. Asymptotes:

   • Horizontal: As $x \to \pm\infty$, $f(x) \to 0$ ⇒ $y=0$

   • No vertical asymptotes

5. First derivative: $f'(x) = \frac{(1)(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$

   Critical points: $f'(x)=0 \Rightarrow 1-x^2=0 \Rightarrow x=\pm 1$

   Sign:

   • $x<-1$: $f' < 0$ (decreasing)

   • $-1<x<1$: $f' > 0$ (increasing)

   • $x>1$: $f' < 0$ (decreasing)

6. Extrema: Local min at $x=-1$ ($f(-1)=-\frac{1}{2}$), local max at $x=1$ ($f(1)=\frac{1}{2}$)

7. Second derivative: $f''(x) = \frac{2x(x^2-3)}{(x^2+1)^3}$

   Inflection points: $x=0$, $x=\pm\sqrt{3}$

8. Sketch: S-shaped curve through origin, max at (1, 1/2), min at (-1, -1/2)

Example 2: Optimization

Find point on parabola $y^2 = 2x$ closest to point (1, 4)

Solution: Distance from (x,y) to (1,4): $D = \sqrt{(x-1)^2 + (y-4)^2}$

Constraint: $y^2 = 2x \Rightarrow x = \frac{y^2}{2}$

Minimize $D^2$ (easier): $f(y) = \left(\frac{y^2}{2}-1\right)^2 + (y-4)^2$

$f'(y) = 2\left(\frac{y^2}{2}-1\right)y + 2(y-4) = y^3 - 2y + 2y - 8 = y^3 - 8$

Set $f'(y)=0$: $y^3=8 \Rightarrow y=2$

Then $x = \frac{y^2}{2} = \frac{4}{2} = 2$

Closest point: (2, 2)

Distance: $D = \sqrt{(2-1)^2 + (2-4)^2} = \sqrt{1+4} = \sqrt{5}$

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16. Common Mistakes and Exam Tips

16.1 Common Mistakes

1. Forgetting to check endpoints in optimization problems

2. Misapplying chain rule in related rates

3. Confusing average rate of change with instantaneous rate

4. Not verifying conditions for Rolle's/MVT

5. Sign errors in derivative calculations

16.2 Problem-Solving Strategy

1. Read carefully: Understand what's given and what's asked

2. Draw diagram: Especially for geometry/optimization problems

3. Define variables: Clearly label all quantities

4. Write equations: Relate variables mathematically

5. Differentiate correctly: Watch for chain rule, product rule, etc.

6. Check units: Ensure consistency in related rates

7. Verify answer: Does it make sense in context?

16.3 Quick Checks

1. Local max/min: $f'$ changes sign at critical point

2. Inflection point: $f''$ changes sign

3. Tangent line: $y - y_0 = f'(x_0)(x - x_0)$

4. Related rates: All derivatives are with respect to same variable (usually time)

5. Optimization: Check both critical points and endpoints

This comprehensive theory covers all applications of derivatives with detailed explanations and examples, providing complete preparation for the entrance examination.