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#### Sarah Thomas

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##
What Is a Derivative Formula?

## What Is a Derivative?

## Elements of a Derivative

## Key Derivative Formulas

### Constant Rule

### Power Rule

### Special Case of the Power Rule

### Constant Multiple Rule

### Chain Rule

### Product Rule

### Quotient Rule

### Sum or Difference Rule

### Trigonometry Rules

### Logarithmic Functions & Exponential Functions Rules

## Solving Derivatives Step By Step

### Step 1

### Step 2

### Step 3

### Practice Example 1

#### Step 1 - Substitute

#### Step 2 - Simplify

#### Step 3 - Evaluate

### Practice Example 2

### Practice Example 3

## Types of Derivatives

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Outlier Articles Home# Derivative Formula: An Easy-To-Understand Guide

## Rachel McLean

Calculus

06.20.2022 • 7 min read

Subject Matter Expert

In this article, we’ll learn derivative formulas. We’ll discuss the limit definition of the derivative and introduce the most common derivative formulas. Finally, we’ll walk through examples of how to find the derivative of a function.

In This Article

Derivative formulas are equations that give quick solutions to common derivative problems. We refer to them as rules—like the power rule and the chain rule, to name a few.

More on these later.

These formulas come from the limit definition of the derivative, and they streamline the differentiation process. That’s why we can also call them differentiation formulas.

The derivative of a function at a point $x$ is equal to the slope of the tangent line at $x$.

This slope value represents the instantaneous rate of change at that point. Differentiation is the process of determining the derivative of a function.

For example, in the graph below, the function $f(x) = \ln{(x)}$ is in blue. The red line is $f(x) = x-1$, which is the line tangent to $f$ at $x = 1$. A tangent line to a point on a function is a line that just barely touches the function at that point. The slope of this tangent line $f(x) = x-1$ is $1$, which means that the derivative of $f(x) = \ln{(x)}$ is $1$ at $x = 1$.

We formally define derivatives using limits:

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

The above equation represents the limit of the average rate of change of $f$ over the interval $[x, x +\Delta{x}]$ as $\Delta{x}$ approaches 0. We also know the average rate of change of a function as the slope of the secant line.

In this notation, $\Delta{x}$ represents a small change in $x$. If this limit exists, then $L$ is the derivative.

The notation $f’(a)$ represents the derivative of a function $f$ at some point $a$. You might hear this notation read aloud as either “the derivative of $f$ evaluated at $a$” or “$f$ prime at $a$.”

The expressions $f’(x)$ and $\frac{dy}{dx}$ both represent the general derivative function of $f$. The latter notation is called Leibniz’s notation. By plugging any point $a$ into the resulting function $f’(x)$, we can determine the slope of the tangent line of $f$ at any point on the curve.

It’s essential to know how to use the limit definition to calculate a derivative. However, the limit definition can be clunky to use. Usually, we rely upon the standard derivative formulas below to differentiate, instead of using the formal limit definition:

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

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

This case is where $n$=1

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

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

$\frac{d}{dx}f(g(x)) = f’(g(x))g’(x)$

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

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

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

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

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

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

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

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

Two game-changing derivative rules, according to Dr. Tim Chartier, are the product rule and quotient rule:

He also explains more differentiation formulas for finding with examples:

Now, we’ll discuss how to find derivatives. We can solve a derivative in two ways:

Using the formal limit derivative definition

Using the basic derivative formulas

Here are the steps to finding derivatives using the limit definition:

Substitute your function into the limit definition of a derivative formula:

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

The hardest part of this step is the correct substitution of $x+h$ in the first term. You need to substitute $x$ with the expression $(x + \Delta{x})$ wherever $x$ appears in $f(x)$.

Simplify.

Evaluate the resulting limit.

For example, let’s find the derivative of the function $f(x) = 3x^2$.

Let's sub in our function $f(x) = 3x^2$ into the limit definition of a derivative.

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

$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3(x + \Delta{x})^2 - 3x^2}{\Delta{x}}$

We can do this by expanding the term $3(x + \Delta{x})^2$ and then combining like terms. Then we can divide by $\Delta{x}$ since $\Delta{x}$ is present in all terms of the numerator and denominator.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3(x^2 + 2x\Delta{x} + \Delta{x}^2) - 3x^2}{\Delta{x}}$

$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{3x^2 + 6x\Delta{x} + 3\Delta{x}^2 - 3x^2}{\Delta{x}}$

$= \mathop {\lim }\limits_{\Delta{x} \to 0} \frac{6x\Delta{x} + 3\Delta{x}^2}{\Delta{x}}$

$= \mathop {\lim }\limits_{\Delta{x} \to 0}6x + 3\Delta{x}$

Now examine the limit as $\Delta{x}$ approaches 0. Polynomials are always continuous. To evaluate this limit, we can substitute $\Delta{x} = 0$ directly into the function we’re left with.

$f’(x)= \mathop {\lim }\limits_{\Delta{x} \to 0}6x + 3\Delta{x}$

$= 6x + 3(0)$

$= 6x$

So, we’ve found that the derivative of $f(x) = 3x^2$ is $f’(x) = 6x$. This is the general derivative formula for any point on the curve of $f$.

To find the derivative at a single point, we can plug $x = a$ into $f’(x) = 6x$.

For example, we can say that:

$f’(1) = 6(1) = 6$, which represents the slope of the tangent line at $x = 1$.

$f’(5) = 6(5) = 30$, which represents the slope of the tangent line at $x = 5$.

$f’(-60) = 6(-60) = -360$, which represents the slope of the tangent line at $x = -60$.

$f’(0) = 6(0) = 0$, which represents the slope of the tangent line at $x = 0$.

We can also use the differentiation formulas to evaluate derivatives. For example, let’s find the derivative of $f(x) = 4x^4 + e^x$.

We can use the sum rule, which states that the derivative of a sum of functions is equal to the sum of their derivatives. We’ll also need to use the power rule for the $4x^4$ term, where our exponent is $n = 4$. Finally, we’ll use the exponential function rule for $e^x$, which tells us that $\frac{d}{dx}e^x = e^x$.

This gives us:

$f’(x) = \frac{d}{dx}4x^4 + \frac{d}{dx}e^x$

$=4 \cdot 4x^{4-1} + e^x$

$= 16x^3 + e^x$

For a slightly trickier example, let’s find the derivative of $f(x) = \sin{(4x)}\cos{(2x)}$.

First, we’ll need the product rule. This says that the derivative of a product of functions is the sum of the first function times the derivative of the second, and the second function times the derivative of the first.

$f’(x) = \sin{(4x)} \cdot \frac{d}{dx}\cos{(2x)} + \cos{(2x)} \cdot \frac{d}{dx}\sin{(4x)}$

Along with the sine and cosine derivative rules, we’ll also need the chain rule, since we have a composition of functions using the sine and cosine functions.

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.

This means that the derivative of $\sin{(4x)}$ is $4\cos{(4x)}$, and the derivative of $\cos{(2x)}$ is $-2\sin{(2x)}$.

$f’(x) = \sin{(4x)} \cdot \frac{d}{dx}\cos{(2x)} + \cos{(2x)} \cdot \frac{d}{dx}\sin{(4x)}$

$= \sin{(4x)} \cdot -2\sin{(2x)} + \cos{(2x)} \cdot 4\cos{(4x)}$

$= -2\sin{(4x)}\sin{(2x)} + 4\cos{(2x)}\cos{(4x)}$

$= 4\cos{(2x)}\cos{(4x)} - 2\sin{(4x)}\sin{(2x)}$

First and second derivatives provide different information about the behavior of a function.

We use the sign of the first derivative to determine if a function is increasing, decreasing, or constant on an interval $I$:

If $f’(x) > 0$ for each $x$ on $I$, then $f$ is increasing on $I$.

If

f’(x) < 0 for eachx onI , thenf is decreasing onI .If $f’(x) = 0$ for each $x$ on $I$, then $f$ is constant on $I$.

We find second derivatives by simply taking the derivative of the first derivative. Second derivatives inform us of the shape of a function. This characteristic is called concavity.

We use the sign of the second derivative to determine intervals of concavity:

If $f’’(x) > 0$ for each $x$ on $I$, then $f$ is concave up on $I$.

If

f’’(x) < 0 for eachx onI , thenf is concave down on $I$.If $f’’(x) = 0$ for each $x$ on $I$, then $f$ has no concavity.

Dr. Hannah Fry discusses first and second derivative tests:

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