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WhizMath

Derivatives: The Mathematics of Change

Introduction: Understanding Rates of Change

Welcome to Whizmath, where we delve into the dynamic world of calculus! Today's lesson introduces one of the most fundamental concepts in mathematics: Derivatives. At its core, a derivative measures how a function changes as its input changes. It's the mathematical tool for understanding rates of change, slopes of curves, and instantaneous velocities.

From predicting the trajectory of a rocket and optimizing production costs in business, to analyzing the speed of chemical reactions and understanding the flow of electricity, derivatives are indispensable across physics, engineering, economics, and countless scientific disciplines. They allow us to move beyond average rates of change to pinpoint the exact rate of change at any given moment.

In this comprehensive lesson, we will define derivatives using limits, explore various rules for differentiation, interpret derivatives graphically, and apply them to real-world problems. Prepare to unlock the power of instantaneous change and gain a deeper insight into how quantities evolve!

Chapter 1: The Concept of a Derivative

To truly understand derivatives, we must first grasp the idea of a limit and how it relates to the slope of a tangent line.

1.1 Average Rate of Change vs. Instantaneous Rate of Change

1.2 Definition of the Derivative (Limit Definition)

The derivative of a function $f(x)$ with respect to $x$, denoted as $f'(x)$ (read "f prime of x") or $\frac{dy}{dx}$, is defined as:

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

This limit represents the slope of the tangent line to the graph of $f(x)$ at any point $x$.

Example 1.2.1: Using the Limit Definition

Find the derivative of $f(x) = x^2$ using the limit definition.

$f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$

$f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h}$

$f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$

$f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h}$

$f'(x) = \lim_{h \to 0} (2x + h)$

$f'(x) = 2x + 0 = 2x$

So, the derivative of $x^2$ is $2x$.

Chapter 2: Basic Differentiation Rules

While the limit definition is fundamental, using it for every function is tedious. Fortunately, we have powerful rules to simplify differentiation.

2.1 Constant Rule

The derivative of a constant function is zero.

$\frac{d}{dx}(c) = 0$

Example: $\frac{d}{dx}(5) = 0$

2.2 Power Rule

If $n$ is any real number, the derivative of $x^n$ is $nx^{n-1}$.

$\frac{d}{dx}(x^n) = nx^{n-1}$

Example: $\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$

Example: $\frac{d}{dx}(\sqrt{x}) = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

2.3 Constant Multiple Rule

The derivative of a constant times a function is the constant times the derivative of the function.

$\frac{d}{dx}(cf(x)) = c \frac{d}{dx}(f(x))$

Example: $\frac{d}{dx}(5x^4) = 5 \frac{d}{dx}(x^4) = 5(4x^3) = 20x^3$

2.4 Sum and Difference Rules

The derivative of a sum or difference of functions is the sum or difference of their derivatives.

$\frac{d}{dx}(f(x) \pm g(x)) = \frac{d}{dx}(f(x)) \pm \frac{d}{dx}(g(x))$

Example: $\frac{d}{dx}(3x^2 + 2x - 7) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x) - \frac{d}{dx}(7) = 6x + 2 - 0 = 6x + 2$

2.5 Product Rule

If $f(x)$ and $g(x)$ are differentiable functions, then the derivative of their product is:

$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$

Example: Find $\frac{d}{dx}((x^2+1)(x^3-2))$

Let $f(x) = x^2+1 \Rightarrow f'(x) = 2x$

Let $g(x) = x^3-2 \Rightarrow g'(x) = 3x^2$

$\frac{d}{dx}((x^2+1)(x^3-2)) = (2x)(x^3-2) + (x^2+1)(3x^2)$

$ = 2x^4 - 4x + 3x^4 + 3x^2$

$ = 5x^4 + 3x^2 - 4x$

2.6 Quotient Rule

If $f(x)$ and $g(x)$ are differentiable functions and $g(x) \neq 0$, then the derivative of their quotient is:

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

Example: Find $\frac{d}{dx}\left(\frac{x^2}{x+1}\right)$

Let $f(x) = x^2 \Rightarrow f'(x) = 2x$

Let $g(x) = x+1 \Rightarrow g'(x) = 1$

$\frac{d}{dx}\left(\frac{x^2}{x+1}\right) = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2}$

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

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

2.7 Chain Rule

Used for differentiating composite functions (a function within a function). If $y = f(g(x))$, then:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

Example: Find $\frac{d}{dx}((2x+3)^5)$

Let $f(u) = u^5$ and $u = g(x) = 2x+3$.

$f'(u) = 5u^4$ and $g'(x) = 2$

$\frac{d}{dx}((2x+3)^5) = 5(2x+3)^4 \cdot 2 = 10(2x+3)^4$

Chapter 3: Derivatives of Common Functions

Beyond polynomial functions, derivatives of trigonometric, exponential, and logarithmic functions are essential.

3.1 Trigonometric Functions

Example: $\frac{d}{dx}(\sin(3x)) = \cos(3x) \cdot \frac{d}{dx}(3x) = 3\cos(3x)$ (using Chain Rule)

3.2 Exponential and Logarithmic Functions

Example: $\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = 2xe^{x^2}$ (using Chain Rule)

Example: $\frac{d}{dx}(\ln(5x)) = \frac{1}{5x} \cdot \frac{d}{dx}(5x) = \frac{1}{5x} \cdot 5 = \frac{1}{x}$

Chapter 4: Applications of Derivatives

Derivatives are powerful tools for solving a wide range of problems involving change and optimization.

4.1 Slope of a Tangent Line

The derivative $f'(x)$ gives the slope of the line tangent to the curve $y=f(x)$ at any point $x$.

Example 4.1.1:

Find the slope of the tangent line to $f(x) = x^3 - 2x$ at $x=2$.

First, find the derivative: $f'(x) = 3x^2 - 2$

Then, evaluate $f'(2)$: $f'(2) = 3(2)^2 - 2 = 3(4) - 2 = 12 - 2 = 10$

The slope of the tangent line at $x=2$ is 10.

4.2 Rates of Change (Physics/Kinematics)

Example 4.2.1:

A ball's height (in meters) is given by $s(t) = -5t^2 + 20t + 1$. Find its velocity at $t=1$ second.

Find the velocity function: $v(t) = s'(t) = \frac{d}{dt}(-5t^2 + 20t + 1) = -10t + 20$

Evaluate $v(1)$: $v(1) = -10(1) + 20 = 10$ m/s

4.3 Optimization (Finding Maxima and Minima)

Derivatives are used to find local maximum and minimum values of a function. These occur at critical points, where $f'(x) = 0$ or $f'(x)$ is undefined.

Steps for Optimization:

  1. Find the first derivative $f'(x)$.
  2. Set $f'(x) = 0$ and solve for $x$ to find critical points.
  3. Use the First Derivative Test (sign changes of $f'(x)$) or Second Derivative Test ($f''(x)$) to classify critical points as local maxima, minima, or neither.

Example 4.3.1:

Find the local extrema of $f(x) = x^3 - 3x^2 + 2$.

1. Find $f'(x)$: $f'(x) = 3x^2 - 6x$

2. Set $f'(x) = 0$: $3x^2 - 6x = 0 \Rightarrow 3x(x - 2) = 0$

   Critical points at $x=0$ and $x=2$.

3. Use First Derivative Test:

   For $x < 0$ (e.g., $x=-1$): $f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 > 0$ (increasing)

   For $0 < x < 2$ (e.g., $x=1$): $f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 < 0$ (decreasing)

   For $x > 2$ (e.g., $x=3$): $f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 > 0$ (increasing)

Local maximum at $x=0$: $f(0) = (0)^3 - 3(0)^2 + 2 = 2$. Point: $(0, 2)$.

Local minimum at $x=2$: $f(2) = (2)^3 - 3(2)^2 + 2 = 8 - 12 + 2 = -2$. Point: $(2, -2)$.

Chapter 5: Higher-Order Derivatives and Implicit Differentiation

Derivatives can be taken multiple times, and we can differentiate equations where $y$ is not explicitly defined as a function of $x$.

5.1 Higher-Order Derivatives

The second derivative, third derivative, and so on are called higher-order derivatives.

Example: If $f(x) = x^4 - 3x^2 + 5x$, find $f''(x)$.

$f'(x) = 4x^3 - 6x + 5$

$f''(x) = \frac{d}{dx}(4x^3 - 6x + 5) = 12x^2 - 6$

5.2 Implicit Differentiation

Used when an equation defines $y$ implicitly as a function of $x$ (e.g., $x^2 + y^2 = 25$). Differentiate both sides of the equation with respect to $x$, remembering to apply the chain rule when differentiating terms involving $y$ (since $y$ is a function of $x$, $\frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx}$).

Example 5.2.1:

Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$.

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)$

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

Isolate $\frac{dy}{dx}$:

$2y \frac{dy}{dx} = -2x$

$\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$

Chapter 6: Common Pitfalls and Tips for Success

Derivatives can be challenging, but being aware of common mistakes can significantly improve your accuracy:

Tips for Success:

Conclusion: Embracing the Dynamics of Calculus

You have now embarked on the exciting journey into differential calculus, gaining a solid understanding of derivatives as the fundamental tool for measuring instantaneous rates of change. From the rigorous limit definition to the efficient application of differentiation rules, you've seen how derivatives allow us to analyze the behavior of functions with unprecedented precision.

The power of derivatives extends far beyond abstract mathematical problems, providing crucial insights into motion, optimization, and the very dynamics of the physical and economic world. As you continue to explore calculus, remember that derivatives are not just about finding slopes; they are about understanding how everything changes.

Keep exploring change, keep calculating rates, and keep mastering calculus with Whizmath!

Practice Problems (with Solutions)

Problem 1: Power Rule

Find the derivative of $f(x) = 5x^3 - 2x^2 + 7x - 10$.

Show Solution

Solution 1:

$f'(x) = \frac{d}{dx}(5x^3) - \frac{d}{dx}(2x^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(10)$

$f'(x) = 5(3x^2) - 2(2x) + 7(1) - 0$

$f'(x) = 15x^2 - 4x + 7$

Problem 2: Product Rule

Find the derivative of $y = (x^2 + 3)(e^x)$.

Show Solution

Solution 2:

Let $f(x) = x^2 + 3 \Rightarrow f'(x) = 2x$

Let $g(x) = e^x \Rightarrow g'(x) = e^x$

Using Product Rule: $y' = f'(x)g(x) + f(x)g'(x)$

$y' = (2x)(e^x) + (x^2 + 3)(e^x)$

$y' = e^x(2x + x^2 + 3)$

Problem 3: Chain Rule

Find the derivative of $f(x) = \sin(4x - \pi)$.

Show Solution

Solution 3:

Let outer function be $\sin(u)$ and inner function $u = 4x - \pi$.

Derivative of outer: $\cos(u)$

Derivative of inner: $4$

Using Chain Rule: $f'(x) = \cos(4x - \pi) \cdot 4$

$f'(x) = 4\cos(4x - \pi)$

Problem 4: Implicit Differentiation

Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$.

Show Solution

Solution 4:

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy)$

$3x^2 + 3y^2 \frac{dy}{dx} = 6(1 \cdot y + x \cdot \frac{dy}{dx})$ (using Product Rule on $6xy$)

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

Group terms with $\frac{dy}{dx}$:

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

Factor out $\frac{dy}{dx}$:

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

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

Problem 5: Optimization

Find the critical points of $f(x) = x^3 - 12x + 1$.

Show Solution

Solution 5:

Find the first derivative: $f'(x) = 3x^2 - 12$

Set $f'(x) = 0$:

$3x^2 - 12 = 0$

$3(x^2 - 4) = 0$

$3(x - 2)(x + 2) = 0$

Critical points: $x = 2, x = -2$