Derivatives Made Simple

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes. Imagine a car’s position changing over time—the derivative tells you its speed at any given moment. In mathematical terms, a derivative is the slope of a function’s tangent line at a specific point. This concept is pivotal in fields like physics (for motion), economics (for cost analysis), and engineering (for optimization). In this guide, we’ll explore derivatives through detailed examples, rules, graphs, and real-world applications, making them intuitive and accessible.

Definition

The derivative of a function $f (x)$ at a point $x$ measures its instantaneous rate of change. It’s defined as the limit of the difference quotient:

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

This formula calculates the slope of the tangent line to the curve at $x$ . If the limit exists, the function is differentiable at that point. Geometrically, as $h$ approaches 0, the secant line (connecting points $(x, f (x))$ and $(x + h, f (x + h))$ ) becomes the tangent line, and its slope is the derivative. In practical terms, the derivative tells us how sensitive a function is to small changes in $x$ .

Examples: $f (x) = x^{2}$ , $x^{3}$ , $\sqrt{x}$ , $\frac{1}{x}$ , $x^{5}$

Let’s compute the derivatives of several functions using the limit definition to build intuition. We’ll also interpret the results to understand what the derivatives tell us about the functions’ behavior.

Example 1: $f (x) = x^{2}$

Let’s find the derivative using the limit definition:

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

= lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} - x^{2}}{h}

= lim_{h \to 0} \frac{2 x h + h^{2}}{h}

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

= 2 x

The derivative is $f^{'} (x) = 2 x$ . This means the slope of the tangent to $y = x^{2}$ at any point $x$ is $2 x$ . For instance:

At $x = 1$ , the slope is $2 \times 1 = 2$ .
At $x = 2$ , the slope is $2 \times 2 = 4$ .
At $x = - 1$ , the slope is $2 \times (- 1) = - 2$ , indicating a downward slope.

This linear increase in slope reflects the parabolic shape of

y = x^{2}

, which steepens as

| x |

grows.

Example 2: $f (x) = x^{3}$

Now for a cubic function:

f^{'} (x) = lim_{h \to 0} \frac{(x + h)^{3} - x^{3}}{h}

= lim_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} - x^{3}}{h}

= lim_{h \to 0} \frac{3 x^{2} h + 3 x h^{2} + h^{3}}{h}

= lim_{h \to 0} (3 x^{2} + 3 x h + h^{2})

= 3 x^{2}

So, $f^{'} (x) = 3 x^{2}$ . The slope grows quadratically:

At $x = 1$ , the slope is $3 \times 1^{2} = 3$ .
At $x = 2$ , the slope is $3 \times 2^{2} = 12$ .
At $x = 0$ , the slope is $0$ , indicating a flat tangent at the origin.

The quadratic growth of the derivative reflects the rapid steepening of

y = x^{3}

Example 3: $f (x) = \sqrt{x}$

For the square root function (where $f (x) = x^{1 / 2}$ ):

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

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

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

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

= lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}

= \frac{1}{2 \sqrt{x}}

Thus, $f^{'} (x) = \frac{1}{2 \sqrt{x}}$ . The slope decreases as $x$ increases:

At $x = 4$ , the slope is $\frac{1}{2 \times 2} = \frac{1}{4}$ .
At $x = 1$ , the slope is $\frac{1}{2 \times 1} = \frac{1}{2}$ .
At $x = 9$ , the slope is $\frac{1}{2 \times 3} = \frac{1}{6}$ .

This decreasing slope mirrors the flattening of

y = \sqrt{x}

x

grows.

Example 4: $f (x) = \frac{1}{x}$

For the function $f (x) = x^{- 1}$ :

f^{'} (x) = lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}

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

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

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

= - \frac{1}{x^{2}}

The derivative is $f^{'} (x) = - \frac{1}{x^{2}}$ . This is always negative, indicating a decreasing function:

At $x = 1$ , the slope is $- \frac{1}{1^{2}} = - 1$ .
At $x = 2$ , the slope is $- \frac{1}{2^{2}} = - \frac{1}{4}$ .
At $x = - 1$ , the slope is $- \frac{1}{(- 1)^{2}} = - 1$ .

The slope’s magnitude decreases with

| x |

, consistent with the hyperbola

y = \frac{1}{x}

Example 5: $f (x) = x^{5}$

For a higher-degree polynomial:

f^{'} (x) = lim_{h \to 0} \frac{(x + h)^{5} - x^{5}}{h}

= lim_{h \to 0} \frac{x^{5} + 5 x^{4} h + 10 x^{3} h^{2} + 10 x^{2} h^{3} + 5 x h^{4} + h^{5} - x^{5}}{h}

= lim_{h \to 0} \frac{5 x^{4} h + 10 x^{3} h^{2} + 10 x^{2} h^{3} + 5 x h^{4} + h^{5}}{h}

= lim_{h \to 0} (5 x^{4} + 10 x^{3} h + 10 x^{2} h^{2} + 5 x h^{3} + h^{4})

= 5 x^{4}

So, $f^{'} (x) = 5 x^{4}$ . The slope grows rapidly:

At $x = 1$ , the slope is $5 \times 1^{4} = 5$ .
At $x = 2$ , the slope is $5 \times 2^{4} = 5 \times 16 = 80$ .
At $x = 0$ , the slope is $0$ , indicating a flat tangent.

This rapid increase reflects the steep nature of

y = x^{5}

x

moves from 0.

Power Rule

The power rule simplifies finding derivatives of power functions, avoiding the limit definition. For a function $f (x) = x^{n}$ , where $n$ is any real number, the derivative is:

f^{'} (x) = n x^{n - 1}

This rule applies to positive, negative, and fractional exponents. Let’s verify with our examples:

$f (x) = x^{2}$ : $n = 2$ , so $f^{'} (x) = 2 x^{2 - 1} = 2 x$ .
$f (x) = x^{3}$ : $n = 3$ , so $f^{'} (x) = 3 x^{3 - 1} = 3 x^{2}$ .
$f (x) = \sqrt{x} = x^{1 / 2}$ : $n = \frac{1}{2}$ , so $f^{'} (x) = \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$ .
$f (x) = \frac{1}{x} = x^{- 1}$ : $n = - 1$ , so $f^{'} (x) = - 1 x^{- 1 - 1} = - 1 x^{- 2} = - \frac{1}{x^{2}}$ .
$f (x) = x^{5}$ : $n = 5$ , so $f^{'} (x) = 5 x^{5 - 1} = 5 x^{4}$ .
Additional example: $f (x) = x^{- 3}$ : $n = - 3$ , so $f^{'} (x) = - 3 x^{- 3 - 1} = - 3 x^{- 4} = - \frac{3}{x^{4}}$ .

The power rule streamlines calculations by leveraging this consistent pattern, making it a cornerstone of calculus.

Graphical View

Graphs provide a visual way to understand derivatives as slopes of tangent lines. Let’s examine three functions to see how their derivatives correspond to their shapes.

For $y = x^{2}$ :

The derivative is $f^{'} (x) = 2 x$ . At $x = 1$ , the slope is 2. The parabola steepens as $| x |$ increases.

For $y = x^{3}$ :

The derivative is $f^{'} (x) = 3 x^{2}$ . At $x = 1$ , the slope is 3. The cubic function’s steepness grows faster than the quadratic.

For $y = \sqrt{x}$ :

The derivative is $f^{'} (x) = \frac{1}{2 \sqrt{x}}$ . At $x = 1$ , the slope is $\frac{1}{2}$ . The curve flattens as $x$ increases.

Applications

Derivatives have wide-ranging applications in science, engineering, economics, and more. Here are some detailed examples:

Physics - Velocity and Acceleration: If a car’s position is $s (t) = 4 t^{2} + 2 t$ , the velocity is $v (t) = s^{'} (t) = 8 t + 2$ . At $t = 3$ seconds, the velocity is $8 \times 3 + 2 = 26$ units/second. The acceleration is $a (t) = v^{'} (t) = 8$ , a constant.
Optimization - Maximum Volume: A box with a square base has volume $V = x^{2} h$ and surface area $x^{2} + 4 x h = 24$ (open top). Solve $h = \frac{24 - x^{2}}{4 x}$ , so $V = x^{2} (\frac{24 - x^{2}}{4 x}) = 6 x - \frac{x^{3}}{4}$ . The derivative is $V^{'} (x) = 6 - \frac{3 x^{2}}{4}$ . Set $V^{'} (x) = 0$ : $6 = \frac{3 x^{2}}{4}$ , so $x^{2} = 8$ , $x = 2 \sqrt{2}$ , maximizing volume.
Economics - Marginal Revenue: If revenue is $R (x) = 50 x - 0.5 x^{2}$ , the marginal revenue is $R^{'} (x) = 50 - x$ . At $x = 20$ , it’s $50 - 20 = 30$ , adding $30 per unit.
Biology - Population Growth: A population grows as $P (t) = 500 e^{0.03 t}$ . The growth rate is $P^{'} (t) = 500 \times 0.03 e^{0.03 t} = 15 e^{0.03 t}$ . At $t = 0$ , it’s 15 individuals per unit time.
Engineering - Beam Deflection: The deflection $y (x) = \frac{w}{24 E I} x^{4}$ has a slope $y^{'} (x) = \frac{w}{6 E I} x^{3}$ , aiding bending analysis.

Derivatives Made Simple

Definition

Examples: f(x)=x2, x3, x, 1x, x5

Example 1: f(x)=x2

Example 2: f(x)=x3

Example 3: f(x)=x

Example 4: f(x)=1x

Example 5: f(x)=x5