Basic Derivative Rules

Learn shortcuts that make differentiation fast and easy

If you have ever watched someone solve a Rubik’s cube in seconds, you know there is a difference between understanding how something works and having efficient techniques to do it quickly. The same is true for derivatives. In the previous lesson, you learned what a derivative is and how to compute it using limits. That process works, but it is slow and tedious. Imagine having to set up and evaluate a limit every single time you wanted to find a derivative. For a polynomial with five terms, you would spend half an hour on what should be a thirty-second problem.

The good news is that mathematicians have already done the hard work for you. They applied the limit definition to common types of functions and discovered patterns—rules that let you skip straight to the answer. These rules are like keyboard shortcuts: once you know them, you will never go back to the long way. By the end of this lesson, you will be able to differentiate polynomials almost as fast as you can write them down.

Core Concepts

Why We Need Rules

Let us revisit the limit definition of the derivative:

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

To find the derivative of something as simple as $f(x) = x^3$ using this definition, you would need to:

  1. Compute $f(x+h) = (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
  2. Subtract $f(x)$ to get $3x^2h + 3xh^2 + h^3$
  3. Divide by $h$ to get $3x^2 + 3xh + h^2$
  4. Take the limit as $h \to 0$ to get $3x^2$

That is a lot of algebra for one term. Now imagine doing that for $f(x) = 4x^5 - 3x^4 + 2x^3 - x^2 + 5x - 7$. You could spend an entire afternoon on it.

The derivative rules we are about to learn are proven once using the limit definition, and then we can use them forever without going through that process again. It is like proving that $2 + 2 = 4$ once and then just using that fact rather than counting on your fingers every time.

The Constant Rule

Think about the graph of a constant function like $f(x) = 5$. It is a horizontal line. What is the slope of a horizontal line? Zero, everywhere. Since the derivative measures slope, the derivative of any constant must be zero.

The Constant Rule: $$\frac{d}{dx}[c] = 0$$

where $c$ is any constant.

This makes intuitive sense. A constant does not change, so its rate of change is zero. The temperature in a perfectly insulated room stays the same, so the rate of temperature change is zero. Your bank balance while you are asleep (hopefully) stays constant, so the rate of change of your wealth is zero.

The Power Rule

This is the workhorse of differential calculus. It handles any function of the form $x^n$ where $n$ is any real number.

The Power Rule: $$\frac{d}{dx}[x^n] = nx^{n-1}$$

In words: bring the exponent down as a coefficient, then reduce the exponent by one.

Let us verify this makes sense with a case you might remember. The derivative of $x^2$ should be $2x$ (the slope of a parabola at any point). Using the power rule: bring down the 2, reduce the exponent by one, and you get $2x^{2-1} = 2x^1 = 2x$. It works.

Here is a simple way to remember it: the exponent “jumps down” to become a multiplier, and as payment for jumping, it decreases by one.

Original Bring Down Exponent Reduce Power Result
$x^5$ $5 \cdot x^{?}$ $5 - 1 = 4$ $5x^4$
$x^{10}$ $10 \cdot x^{?}$ $10 - 1 = 9$ $10x^9$
$x^1$ $1 \cdot x^{?}$ $1 - 1 = 0$ $1x^0 = 1$

Notice that last row: the derivative of $x$ is just $1$. The graph of $y = x$ is a line with slope 1, so this checks out.

The Constant Multiple Rule

What if you have $3x^4$ instead of just $x^4$? You might guess that you can just “pull out” the constant and deal with it separately. You would be right.

The Constant Multiple Rule: $$\frac{d}{dx}[c \cdot f(x)] = c \cdot f’(x)$$

where $c$ is any constant.

In words: constants pass through the derivative operation unchanged. You can differentiate the function part and keep the constant multiplier.

So for $3x^4$: $$\frac{d}{dx}[3x^4] = 3 \cdot \frac{d}{dx}[x^4] = 3 \cdot 4x^3 = 12x^3$$

This rule reflects a real-world truth. If you are driving three times as fast as your friend, your rate of change of position (velocity) is three times theirs at every moment. Scaling a function scales its derivative by the same factor.

The Sum and Difference Rules

What about $f(x) = x^3 + x^2$? Can you just take the derivative of each piece and add them together? Yes, you can.

The Sum Rule: $$\frac{d}{dx}[f(x) + g(x)] = f’(x) + g’(x)$$

The Difference Rule: $$\frac{d}{dx}[f(x) - g(x)] = f’(x) - g’(x)$$

In words: the derivative of a sum is the sum of the derivatives, and the derivative of a difference is the difference of the derivatives.

Think of it this way. If you have two bank accounts, and one is growing at $100 per month while the other is growing at $50 per month, your total wealth is growing at $150 per month. The rates of change add up.

Combining the Rules: Polynomials

Now you have all the tools you need to differentiate any polynomial almost instantly. A polynomial is just a sum of constant multiples of powers of $x$, and you know how to handle all of those pieces.

Let us differentiate $f(x) = 4x^5 - 3x^3 + 2x - 7$:

$$\frac{d}{dx}[4x^5 - 3x^3 + 2x - 7]$$

Using the sum and difference rules, take the derivative of each term:

$$= \frac{d}{dx}[4x^5] - \frac{d}{dx}[3x^3] + \frac{d}{dx}[2x] - \frac{d}{dx}[7]$$

Using the constant multiple rule and constant rule:

$$= 4\frac{d}{dx}[x^5] - 3\frac{d}{dx}[x^3] + 2\frac{d}{dx}[x] - 0$$

Using the power rule:

$$= 4(5x^4) - 3(3x^2) + 2(1)$$

$$= 20x^4 - 9x^2 + 2$$

With practice, you will do all of this in your head and write down the answer directly.

Extending the Power Rule: Negative Exponents

The power rule works for any real exponent, not just positive integers. This is incredibly useful because it lets you differentiate functions involving division by powers of $x$.

Remember that $\frac{1}{x^n} = x^{-n}$. So if you need to differentiate $\frac{1}{x^3}$, you can rewrite it as $x^{-3}$ and apply the power rule:

$$\frac{d}{dx}[x^{-3}] = -3x^{-3-1} = -3x^{-4} = \frac{-3}{x^4}$$

The rule still works the same way: bring down the exponent (which happens to be negative), and reduce it by one (making it more negative).

Extending the Power Rule: Fractional Exponents

Roots can be written as fractional exponents:

  • $\sqrt{x} = x^{1/2}$
  • $\sqrt[3]{x} = x^{1/3}$
  • $\sqrt[n]{x} = x^{1/n}$

This means the power rule handles all roots too.

For example, to differentiate $\sqrt{x}$:

$$\frac{d}{dx}[\sqrt{x}] = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$

You can always convert back to radical notation at the end if you prefer, but the power rule requires the exponential form to apply.

Higher-Order Derivatives

Since the derivative of a function is itself a function, you can take its derivative too. The derivative of the derivative is called the second derivative.

Notation for higher derivatives:

Order Prime Notation Leibniz Notation
First $f’(x)$ $\frac{dy}{dx}$ or $\frac{df}{dx}$
Second $f’’(x)$ $\frac{d^2y}{dx^2}$
Third $f’’’(x)$ $\frac{d^3y}{dx^3}$
Fourth $f^{(4)}(x)$ $\frac{d^4y}{dx^4}$

After the third derivative, we switch to the superscript number notation because writing four or more primes gets messy.

To find a second derivative, you simply differentiate twice:

If $f(x) = x^4$, then:

  • $f’(x) = 4x^3$ (first derivative)
  • $f’’(x) = 12x^2$ (second derivative)
  • $f’’’(x) = 24x$ (third derivative)
  • $f^{(4)}(x) = 24$ (fourth derivative)
  • $f^{(5)}(x) = 0$ (fifth derivative and all higher ones)

Notice that for a polynomial of degree $n$, the $(n+1)$th derivative and all subsequent derivatives are zero.

Notation and Terminology

Term Meaning Example
Power rule $\frac{d}{dx}[x^n] = nx^{n-1}$ $\frac{d}{dx}[x^5] = 5x^4$
Constant rule $\frac{d}{dx}[c] = 0$ $\frac{d}{dx}[7] = 0$
Constant multiple rule $\frac{d}{dx}[cf(x)] = cf’(x)$ $\frac{d}{dx}[3x^2] = 6x$
Sum rule Derivative of sum equals sum of derivatives $\frac{d}{dx}[x^2 + x] = 2x + 1$
Second derivative Derivative of the derivative $f’’(x)$, measures acceleration
$\frac{d^2y}{dx^2}$ Leibniz notation for second derivative Read as “d two y, d x squared”
$f^{(n)}(x)$ The $n$th derivative of $f$ $f^{(4)}(x)$ is the fourth derivative
Differentiate The act of finding a derivative “Differentiate $x^3$” means find $\frac{d}{dx}[x^3]$

Examples

Example 1: Basic Power Rule

Problem: Find $\frac{d}{dx}[x^7]$

Solution:

Using the power rule, bring down the exponent and reduce it by one:

$$\frac{d}{dx}[x^7] = 7x^{7-1} = 7x^6$$

The derivative of $x^7$ is $7x^6$.

Example 2: Polynomial with Multiple Terms

Problem: Find $\frac{d}{dx}[5x^3 - 2x + 7]$

Solution:

Apply the sum and difference rules to differentiate each term separately:

$$\frac{d}{dx}[5x^3 - 2x + 7] = \frac{d}{dx}[5x^3] - \frac{d}{dx}[2x] + \frac{d}{dx}[7]$$

Now apply the constant multiple rule and power rule to each term:

  • $\frac{d}{dx}[5x^3] = 5 \cdot 3x^2 = 15x^2$
  • $\frac{d}{dx}[2x] = 2 \cdot 1 = 2$
  • $\frac{d}{dx}[7] = 0$

Combining:

$$\frac{d}{dx}[5x^3 - 2x + 7] = 15x^2 - 2$$

Example 3: Square Root (Fractional Exponent)

Problem: Find $\frac{d}{dx}[\sqrt{x}]$

Solution:

First, rewrite the square root using exponential notation:

$$\sqrt{x} = x^{1/2}$$

Now apply the power rule:

$$\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2}$$

Converting back to radical form:

$$\frac{1}{2}x^{-1/2} = \frac{1}{2} \cdot \frac{1}{x^{1/2}} = \frac{1}{2\sqrt{x}}$$

Therefore, $\frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}}$.

This tells us that the slope of the curve $y = \sqrt{x}$ decreases as $x$ increases, which matches what we see when we graph it—the curve flattens out as $x$ gets larger.

Example 4: Second Derivative

Problem: Find $f’’(x)$ for $f(x) = x^4 - 3x^2 + x$

Solution:

First, find the first derivative $f’(x)$:

$$f’(x) = \frac{d}{dx}[x^4 - 3x^2 + x]$$

$$f’(x) = 4x^3 - 6x + 1$$

Now differentiate again to find the second derivative $f’’(x)$:

$$f’’(x) = \frac{d}{dx}[4x^3 - 6x + 1]$$

$$f’’(x) = 12x^2 - 6$$

This second derivative tells us about the concavity of the original function. When $f’’(x) > 0$, the graph is concave up (curves upward like a smile). When $f’’(x) < 0$, the graph is concave down.

Setting $f’’(x) = 0$: $12x^2 - 6 = 0$, which gives $x = \pm\frac{1}{\sqrt{2}}$. These are the inflection points where concavity changes.

Example 5: Negative and Fractional Exponents Combined

Problem: Find the derivative of $f(x) = \frac{3}{x^2} + 2\sqrt[3]{x}$

Solution:

First, rewrite everything using exponential notation:

$$f(x) = 3x^{-2} + 2x^{1/3}$$

Now apply the sum rule and differentiate each term:

For the first term: $$\frac{d}{dx}[3x^{-2}] = 3 \cdot (-2)x^{-2-1} = -6x^{-3}$$

For the second term: $$\frac{d}{dx}[2x^{1/3}] = 2 \cdot \frac{1}{3}x^{1/3 - 1} = \frac{2}{3}x^{-2/3}$$

Combining: $$f’(x) = -6x^{-3} + \frac{2}{3}x^{-2/3}$$

Converting to radical and fraction form: $$f’(x) = \frac{-6}{x^3} + \frac{2}{3x^{2/3}} = \frac{-6}{x^3} + \frac{2}{3\sqrt[3]{x^2}}$$

Both forms are correct. The exponential form is often easier to work with for further differentiation.

Example 6: Third Derivative

Problem: Find $f’’’(x)$ for $f(x) = 2x^5 - x^4 + 3x^2$

Solution:

We need to differentiate three times.

First derivative: $$f’(x) = 10x^4 - 4x^3 + 6x$$

Second derivative: $$f’’(x) = 40x^3 - 12x^2 + 6$$

Third derivative: $$f’’’(x) = 120x^2 - 24x$$

Notice how each differentiation reduces the degree of each term by one. The constant term $6$ in $f’’(x)$ disappeared when we took the third derivative.

Example 7: Special Cases

Problem: Find the derivatives of: (a) $f(x) = x$, (b) $g(x) = 1$, (c) $h(x) = x^0$

Solution:

(a) For $f(x) = x = x^1$: $$f’(x) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1$$

(b) For $g(x) = 1$ (a constant): $$g’(x) = 0$$

(c) For $h(x) = x^0 = 1$ (also a constant for $x \neq 0$): $$h’(x) = 0$$

These special cases are worth remembering. The derivative of $x$ is always 1, and the derivative of any constant is always 0.

Key Properties and Rules

Here is a complete summary of all the basic derivative rules in one place:

Constant Rule: $$\frac{d}{dx}[c] = 0$$

Power Rule: $$\frac{d}{dx}[x^n] = nx^{n-1} \quad \text{(for any real number } n \text{)}$$

Constant Multiple Rule: $$\frac{d}{dx}[c \cdot f(x)] = c \cdot f’(x)$$

Sum Rule: $$\frac{d}{dx}[f(x) + g(x)] = f’(x) + g’(x)$$

Difference Rule: $$\frac{d}{dx}[f(x) - g(x)] = f’(x) - g’(x)$$

Combined Rule for Polynomials: For $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$: $$f’(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1$$

Higher Derivatives:

  • The second derivative is $f’’(x) = \frac{d}{dx}[f’(x)]$
  • The $n$th derivative is the derivative of the $(n-1)$th derivative
  • For a polynomial of degree $n$, the $(n+1)$th derivative is 0

Important Special Cases:

  • $\frac{d}{dx}[x] = 1$
  • $\frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}}$
  • $\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$
  • $\frac{d}{dx}\left[\frac{1}{x^2}\right] = -\frac{2}{x^3}$

Real-World Applications

Position, Velocity, and Acceleration

The most natural application of higher derivatives comes from physics. If $s(t)$ represents the position of an object at time $t$:

  • Velocity is the first derivative: $v(t) = s’(t)$
  • Acceleration is the second derivative: $a(t) = s’’(t) = v’(t)$

For example, if a ball is thrown upward and its height in meters is given by: $$s(t) = -4.9t^2 + 20t + 1.5$$

Then its velocity is: $$v(t) = s’(t) = -9.8t + 20$$

And its acceleration is: $$a(t) = s’’(t) = -9.8$$

The constant acceleration of $-9.8 , m/s^2$ is gravity. This is why all objects near Earth’s surface accelerate downward at the same rate, regardless of their velocity.

Jerk: The Third Derivative

If acceleration is the rate of change of velocity, what is the rate of change of acceleration? It is called jerk, and it is the third derivative of position.

$$j(t) = s’’’(t) = a’(t)$$

Jerk measures how suddenly acceleration changes. When you are in a car and the driver slams on the brakes, you feel high jerk—a rapid change in acceleration. Smooth elevator rides are designed to minimize jerk so passengers stay comfortable.

In the ball example above, $j(t) = 0$ because gravity provides constant acceleration. But for a rocket whose engines are throttling up, jerk would be nonzero.

Economics: Marginal Analysis

In economics, if $C(x)$ represents the total cost of producing $x$ units:

  • Marginal cost is $C’(x)$, the cost of producing one more unit
  • Rate of change of marginal cost is $C’’(x)$

If $C’’(x) > 0$, marginal costs are increasing—each additional unit costs more than the last. This indicates diminishing returns or capacity constraints.

If $C’’(x) < 0$, marginal costs are decreasing—economies of scale are taking effect.

Concavity and Inflection Points

The second derivative tells us about the shape of a curve:

  • When $f’’(x) > 0$: The graph is concave up (opens upward, like a cup that holds water)
  • When $f’’(x) < 0$: The graph is concave down (opens downward, like an umbrella)
  • When $f’’(x) = 0$: Possible inflection point where concavity changes

For the function $f(x) = x^3$:

  • $f’(x) = 3x^2$
  • $f’’(x) = 6x$

At $x = 0$, $f’’(0) = 0$. For $x < 0$, $f’’(x) < 0$ (concave down). For $x > 0$, $f’’(x) > 0$ (concave up). So $x = 0$ is an inflection point where the graph changes from curving one way to curving the other.

Population Growth Models

Simple population models often use polynomial approximations. If $P(t)$ represents population at time $t$:

  • $P’(t)$ is the rate of population change (births minus deaths)
  • $P’’(t)$ indicates whether growth is accelerating or decelerating

If $P’’(t) > 0$, the population is not just growing—it is growing faster and faster.

Self-Test Problems

Test your understanding with these problems. Try to solve them before looking at the answers.

Problem 1: Find the derivative of $f(x) = 6x^4 - 2x^3 + 5x - 9$.

Show Answer

$$f’(x) = 24x^3 - 6x^2 + 5$$

Explanation: Apply the power rule to each term:

  • $\frac{d}{dx}[6x^4] = 6 \cdot 4x^3 = 24x^3$
  • $\frac{d}{dx}[-2x^3] = -2 \cdot 3x^2 = -6x^2$
  • $\frac{d}{dx}[5x] = 5$
  • $\frac{d}{dx}[-9] = 0$

Problem 2: Find the derivative of $g(x) = \sqrt[4]{x}$.

Show Answer

$$g’(x) = \frac{1}{4}x^{-3/4} = \frac{1}{4\sqrt[4]{x^3}}$$

Explanation: Rewrite $\sqrt[4]{x} = x^{1/4}$, then apply the power rule: $$\frac{d}{dx}[x^{1/4}] = \frac{1}{4}x^{1/4 - 1} = \frac{1}{4}x^{-3/4}$$

Problem 3: Find the derivative of $h(x) = \frac{5}{x^3} - \frac{2}{x}$.

Show Answer

$$h’(x) = \frac{-15}{x^4} + \frac{2}{x^2}$$

Explanation: Rewrite using negative exponents: $$h(x) = 5x^{-3} - 2x^{-1}$$

Apply the power rule: $$h’(x) = 5 \cdot (-3)x^{-4} - 2 \cdot (-1)x^{-2} = -15x^{-4} + 2x^{-2}$$

Converting back: $h’(x) = \frac{-15}{x^4} + \frac{2}{x^2}$

Problem 4: Find the second derivative of $f(x) = x^5 - 4x^3 + 2x$.

Show Answer

$$f’’(x) = 20x^3 - 24x$$

Explanation:

First derivative: $$f’(x) = 5x^4 - 12x^2 + 2$$

Second derivative: $$f’’(x) = 20x^3 - 24x$$

Problem 5: The height of a ball thrown upward is given by $h(t) = -5t^2 + 30t + 2$ meters, where $t$ is in seconds. (a) Find the velocity function $v(t)$. (b) Find the acceleration function $a(t)$. (c) At what time does the ball reach its maximum height?

Show Answer

(a) $v(t) = h’(t) = -10t + 30$ m/s

(b) $a(t) = h’’(t) = -10$ m/s$^2$

(c) Maximum height occurs when velocity equals zero: $$-10t + 30 = 0$$ $$t = 3 \text{ seconds}$$

Explanation: At maximum height, the ball momentarily stops before falling back down, so its velocity is zero. The constant negative acceleration represents gravity pulling the ball downward.

Problem 6: Find the third derivative of $f(x) = 2x^6 - x^4 + 3x^2 - 1$.

Show Answer

$$f’’’(x) = 240x^3 - 24x$$

Explanation:

First derivative: $f’(x) = 12x^5 - 4x^3 + 6x$

Second derivative: $f’’(x) = 60x^4 - 12x^2 + 6$

Third derivative: $f’’’(x) = 240x^3 - 24x$

Problem 7: Find the derivative of $f(x) = 3\sqrt{x} + \frac{4}{\sqrt{x}}$.

Show Answer

$$f’(x) = \frac{3}{2\sqrt{x}} - \frac{2}{x\sqrt{x}} = \frac{3}{2}x^{-1/2} - 2x^{-3/2}$$

Explanation: Rewrite using exponents: $$f(x) = 3x^{1/2} + 4x^{-1/2}$$

Apply the power rule: $$f’(x) = 3 \cdot \frac{1}{2}x^{-1/2} + 4 \cdot \left(-\frac{1}{2}\right)x^{-3/2}$$ $$f’(x) = \frac{3}{2}x^{-1/2} - 2x^{-3/2}$$

In radical form: $f’(x) = \frac{3}{2\sqrt{x}} - \frac{2}{x^{3/2}} = \frac{3}{2\sqrt{x}} - \frac{2}{x\sqrt{x}}$

Problem 8: For what value of $x$ does $f(x) = x^3 - 6x^2 + 9x + 1$ have a horizontal tangent line?

Show Answer

$x = 1$ and $x = 3$

Explanation: A horizontal tangent line has slope zero, so we need $f’(x) = 0$.

$$f’(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$$

Setting $f’(x) = 0$: $$3(x-1)(x-3) = 0$$ $$x = 1 \text{ or } x = 3$$

Summary

In this lesson, you learned the fundamental rules that make differentiation fast and practical:

  • The Constant Rule states that $\frac{d}{dx}[c] = 0$ because constants do not change.

  • The Power Rule states that $\frac{d}{dx}[x^n] = nx^{n-1}$. This works for all real exponents, including negative numbers and fractions.

  • The Constant Multiple Rule allows you to “pull out” constants: $\frac{d}{dx}[cf(x)] = cf’(x)$.

  • The Sum and Difference Rules let you differentiate term by term: the derivative of a sum is the sum of the derivatives.

  • Fractional exponents let you differentiate roots. Remember that $\sqrt{x} = x^{1/2}$ and $\sqrt[n]{x} = x^{1/n}$.

  • Negative exponents let you differentiate expressions with $x$ in the denominator. Remember that $\frac{1}{x^n} = x^{-n}$.

  • Higher-order derivatives are found by differentiating multiple times. The second derivative $f’’(x)$ measures concavity and, in physics, acceleration.

  • Practical applications include analyzing motion (velocity, acceleration, jerk), economics (marginal analysis), and understanding the shape of curves (concavity, inflection points).

These rules form the foundation for all the differentiation you will do in calculus. In the coming lessons, you will learn additional rules for products, quotients, and compositions of functions, which will complete your differentiation toolkit. For now, practice these basic rules until they become automatic—most derivative problems you encounter will use them in some form.