Calculus

# What Is a Derivative in Calculus?

## Rachel McLean

Subject Matter Expert

In this article, we’ll discuss the meaning of slope, tangent, and the derivative. We’ll learn how to derive the limit definition of the derivative. Then, we’ll examine the most common derivative rules and practice with some examples.

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Differentiation is one of the most fundamental operations in calculus. Knowing how to find the derivative of a function will open up many doors in calculus.

## What Is a Derivative?

Derivatives measure rates of change. More specifically, derivatives measure instantaneous rates of change at a point. The instantaneous rate of change of the function at a point is equal to the slope of the tangent line at that point.

The first derivative of a function $f$ at some given point $a$ is denoted by $f’(a)$. This expression is read aloud as “the derivative of $f$ evaluated at $a$” or “$f$ prime at $a$.”

The expression $f’(x)$ is the notation used to denote the general derivative function of $f$. We can also use Leibniz’s notation $\frac{dy}{dx}$ to denote the derivative function. We can plug $x = a$ into $f’(x)$ to determine the instantaneous rate of change of $f$ with respect to $x$ at any point $a$ on the curve.

The derivative of $f$ at $x$ equals the limit of the average rate of change of $f$ over the interval $[x, x +\Delta{x}]$ as $\Delta{x}$ approaches 0, where $\Delta{x}$ represents a change in $x$:

$f’(x) = \mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }=L$

Here are the 3 steps to calculate a derivative using this definition:

1. Substitute your function into the limit definition of a derivative formula.

2. Simplify.

3. Evaluate the resulting limit.

### Example of How To Calculate a Derivative

Let’s do a very simple example together. Find the derivative of $f(x) = 3x$ using the limit definition and the steps given above.

#### Step 1

The first step is to substitute $f(x) = 3x$ into the limit definition of a derivative. The trick to this step is to substitute the variable $x$ with the expression $(x + \Delta{x})$ wherever $x$ appears in $f(x) = 3x$.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x} }$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3(x + \Delta{x}) - 3x}{\Delta{x}}$

### Step 2

The next step is to simplify. We’ll do this by expanding the numerator and combining like terms. Then, we can divide by $\Delta{x}$.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3(x + \Delta{x}) - 3x}{\Delta{x}}$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3x + 3\Delta{x} - 3x}{\Delta{x}}$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3\Delta{x}}{\Delta{x}}$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} 3$

#### Step 3

Finally, we can evaluate the limit.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} 3$
$f’(x)= 3$

Using these steps, we’ve shown that the derivative function of $f(x) = 3x$ is 3.

Since this is a constant function, the derivative of $f(x)$ at any point is 3.

## What Are Slope and Tangent?

In the above section, we learned the limit definition of a derivative:

$f’(x) =\mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }=L$

But what does this equation mean, and how is it derived? We’ll answer this question in this section.

The slope of a straight line through $(a, f(a))$ and $(b, f(b))$ is equal to the change in $y$ divided by the change in $x$:

$\text{Slope } = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b-a}$

But what about curves that aren’t straight? For non-linear functions, we can instead find the slope of the curve at a specific point. The slope of a line at a specific point on a curve is called the slope of the tangent line. This value is equal to the instantaneous rate of change, or derivative, at that point.

The tangent line to a function at a specific point is a line that just barely touches the function at that point.

Take a look at the graph below, where the tangent line to the red curve $f(x) = -\ln{x}$ at $(1,0)$ is already graphed for us. The blue line represents this tangent line and has the equation $f(x) = -x + 1$. The blue line just barely touches the red line at the point $(1, 0)$.

Observe that the tangent line $f(x) = -x + 1$ is given in slope-intercept form. Slope-intercept form is $f(x) = mx +b$, where $m$ is the slope. So, the slope of the tangent line $f(x) = -x + 1$ is -1. Since the derivative of a function at a point is equal to the slope of the tangent line at that point, this tells us that the derivative of $f(x) = \ln{(x)}$ at the point $(1, 0)$ is -1. We can also write $f’(1) = -1$.

Now we’ve learned that the slope of a curve at a specific point on a curve is called the slope of the tangent line. There’s another type of slope that is helpful to us when defining derivatives: the slope of the secant line. The slope between two separate points on a curve is called the slope of the secant line, which is also called the average rate of change.

You’re already familiar with the equation for this; it’s the same as the formula for the slope of a straight line!

$\text{Average Rate of Change} = \frac{\Delta{y}}{\Delta{x}} = \frac{f(b)-f(a)}{b-a}$

Over the interval $[x, x +\Delta{x}]$, this equation is equal to:

$\text{Average Rate of Change} = \frac{f(x + \Delta{x})-f(x)}{\Delta{x}}$

If we make $\Delta{x}$ become closer and closer to 0 in the above equation of the secant line, we get closer and closer to finding the instantaneous rate of change of a function at $x$.

So, to find the instantaneous rate of change of a function at $x$, we can take the limit of the average rate of change of $f$ over the interval $[x, x +\Delta{x}]$ as $\Delta{x}$ approaches 0. This limit is the formal derivative definition formula:

$f’(x) =\mathop{\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x\right)}}{\Delta{x} }=L$

If $L$ exists, then $f$ is differentiable and $L$ is the derivative of the function $f$ at $x$.

We can compute higher order derivatives as well. If the first derivative of a function is differentiable, we can also determine its second derivative. The second derivative is simply the derivative of the first derivative.

## What Are the Derivative Formulas?

Once you understand the definition of a derivative, you can begin to become familiar with the most common derivative formulas. While it’s important to understand how to derive the limit definition of a derivative, these rules offer you some shortcuts to computing derivatives. These formulas allow you to calculate derivatives much faster than using the limit definition of a derivative.

Here are the most frequently used derivative equations. If you forget any of these rules during an exam, you can always rely on the limit definition to calculate the derivative.

### Constant Rule

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

### Power Rule

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

Special Case of the Power Rule (where n=1): $\frac d{dx}(x)=1$

### Constant Multiple Rule

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

### Chain Rule or Composite Functions Rule

$\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$

### Product Rule

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

### Quotient Rule

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

### Sum or Difference Rule

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

### Trigonometry Rules

$\frac{d}{dx}(\sin{(x)}) = \cos{(x)}$

$\frac{d}{dx}(\cos{(x)}) = -\sin{(x)}$

$\frac{d}{dx}(\tan{(x)}) = \sec ^2 (x)$

### Rules for Logarithmic Functions and Exponential Functions

$\frac{d}{dx} (\ln{x}) = \frac{1}{x}$

$\frac{d}{dx}(e^x) = e^x$

Dr. Tim Chartier highlights two of these rules, the Product Rule and Quotient Rule, as game changers:

He also goes over in this video the Constant Rule, Power Rule, and Sum Rule with examples:

## Examples of How To Find the Derivative of a Function

Let’s try a few derivative examples together.

### Finding a Derivative Example 1

Find the derivative of $f(x) = 4x^2$ using the limit definition of a derivative.

### Solution

We’ll follow the three steps listed in the first section.

#### Step 1

Substituting our function $f(x) = 4x^2$ into the limit definition of a derivative, we get:

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{{f\left( {x + \Delta{x} } \right) - f\left( x \right)}}{\Delta{x} }$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{4(x + \Delta{x})^2 - 4x^2}{\Delta{x}}$

#### Step 2

The next step is to simplify. We’ll do this by expanding the numerator and combining like terms. Then, we can divide by $\Delta{x}$.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{4(x^2 + 2x\Delta{x} + \Delta{x}^2) - 4x^2}{\Delta{x}}$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{4x^2 + 8x\Delta{x} + 4\Delta{x}^2 - 4x^2}{\Delta{x}}$
$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{8x\Delta{x} + 4\Delta{x}^2}{\Delta{x}}$
$= \mathop {\lim }\limits_{\Delta{x} \to 0}8x + 4\Delta{x}$

#### Step 3

Now, we can evaluate the limit as $\Delta{x}$ approaches 0. Polynomial functions are always continuous, so we can substitute $\Delta{x} = 0$ into the function.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0}8x + 4\Delta{x}$
$f’(x) = 8x + 4(0)$
$f’(x) = 8x$

### Finding a Derivative Example 2

Let $f(x) = 7x^3 - \sin{(3x)}$. Find $f’(x)$ using the derivative rules.

### Solution

We have the difference of two terms, so we can use the Difference Rule, which states that the derivative of a difference of functions is equal to the difference of their derivatives. To find the derivative of the first term, we can use the Power Rule and the Constant Multiple Rule:

$\frac{d}{dx}[7x^3] = 3 \cdot 7 x^{3-1} = 21x^2$

The second term is a composition of function. So, to find the derivative of the second term, we can use the Chain Rule. We’ll also need to use the sine rule of the trigonometry rules. The Chain Rule states that the derivative of a composition of functions is equal to the derivative of the outside function, multiplied by the derivative of the inside function:

$\frac{dy}{dx} \sin{(3x)} = \cos{(3x)} \cdot 3 = 3\cos{(3x)}$

Now, we can take the difference of these two derivatives to find the derivative of $f(x) = 7x^3 - \sin{(3x)}$:

$f’(x) = 21x^2 - 3\cos{(3x)}$

### Finding a Derivative Example 3

Let $f(x) = \frac{\cos{(2x)}}{2x}$. Find $f’(x)$ using the derivative rules.

### Solution

We have a quotient of functions. So, we can use the Quotient Rule. In the numerator, we’ll need the cosine rule of the trigonometry rules, as well as the Chain Rule. In the denominator, we’ll use the Power Rule.

$\frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x)f’(x)-f(x)g’(x)}{(g(x))^2}$
$f’(x) = \frac{2x \cdot [-\sin{(2x)} \cdot 2] - \cos{(2x)} \cdot 2}{(2x)^2}$
$f’(x) = \frac{-4x\sin{(2x)} - 2\cos{(2x)}}{4x^2}$
$f’(x) = \frac{-\sin{(2x)}}{x} - \frac{\cos{(2x)}}{2x^2}$

## Where To Practice Derivatives

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