Calculus is essential in all branches of mathematics, science, and engineering, as well as in businesses and health-related fields. One of the most important tools in calculus is the **derivative.**

A derivative is known as the instantaneous rate of change of a quantity y with respect to another quantity *x*. The process of finding derivatives is called **differentiation.** A derivative is also defined as the slope of a curve’s tangent at a point.

In this article, we will learn how to calculate derivatives in convenient ways using the several theorems known as the fundamental rules of derivatives.

## Fundamental Rules of Derivatives

Recall that the definition of derivative is:

Given any number x for which the limit

f’ (x) = $\frac{f(x+h)-f(x)}{h}$

exists, we assign to x the number f’ (x).

Differentiation using this definition is quite tedious in finding the derivative of a function. So, we will explore several shortcuts that can be used to compute the derivative without having to use the limit definition directly.

In this section, we will be discussing the fundamental rules of derivatives along with their proofs.

**The Constant Rule**

We know that the graph of a constant function is a horizontal line. And the rate of change or the slope of a constant function is 0. Similarly, the constant rule states that the derivative of a constant function is zero.

Let c be a constant.

If f(x)=c, then f'(x)=0.

Alternatively, we can state this rule as

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

**Proof**

To prove the constant rule, let us apply the limit definition of derivatives in finding the derivative of the constant function, f(x)=c.

Suppose both f(x)=c and f(x+h)=c. So, we obtain the following result:

f’ (x) = $\frac{f(x+h)-f(x)}{h}$

= $\frac{c-c}{h}$

= $\frac{0}{h}$

= 0=0

**Example**

What is the derivative of f (x)= 3?

**Solution**

Since f (x)= 3 is a constant function then, by the constant rule of derivatives,

f'(x)= = 0**.**

**The Power Rule**

The power rule tells us that if a function is an algebraic expression raised to a power, then the derivative has a power of 1 less than the original function.

Let n be a positive integer if f (x)= x^{n}, then f’ (x) = nx^{n-1}. Alternatively, we can state this rule as $\frac{d}{dx}$ x^{n}= nx^{n-1}.

However, you must remember that the power rule is not applicable to functions with a constant raised to a variable power, such as f (x)=3^{x}.

**Proof**

For this proof, we need to restrict n to be a positive integer. If we define f(x)= x^{n}, we know that by the limit definition of derivative f’ (a) would be

$f’ (a) = \frac{f(x)-f(a)}{x-a} = \frac{x^n – a^n}{x-a}$

We take x^{n} – a^{n} and factor it so we can have

x^{n} – a^{n} = (x – a)(x^{n-1} + x^{n-2} a + ⋯ + xa^{n-2} + a^{n-1})

Notice that the second factor has n numbers of terms. Now, if we substitute this to our above function, we’ll have

f’ (a) = $\frac{(x – a)(x^{n-1} + x^{n-2} a + ⋯ + xa^{n-2} + a^{n-1})}{x-a}$

We can cancel the x-a in the numerator and denominator and compute for the limit

f’ (a)= ( x^{n-1} + x^{n-2} a + ⋯ + xa^{n-2} + a^{n-1} )

f’ (a)= ( a^{n-1} + a^{n-2} a + ⋯ + aa^{n-2} + a^{n-1})

We simplify the right-hand side and observe that all the terms will be the same. And since we know that there are n number of terms in the second factor, we obtain

f’ (a)= na^{n-1}

To completely finish the proof, we simply substitute a with x to get,

f’ (x)= nx^{n-1}

**Example #1**

What is the derivative of f (x)= x^{3} ?

**Solution**

f’ (x) =(3)x^{3-1}

f’ (x) =3x^{2}

**Example #2**

What is the derivative of f (x)=2x^{5}?

**Solution**

f’ (x) =2(5)x^{5-1}

f’ (x) =10x^{4}

The power rule holds for any real number n. However, the proof for the general case, where n is a nonpositive integer, is a bit more complicated, so we will not proceed with it.

**The Sum and Difference, and Constant Multiple Rule**

Working with the derivative of multiple functions, such as finding their sum and differences or multiplying a function with a constant, can be made easier with the following rules.

Let f (x) and g (x) be differentiable functions and let k be a constant. Then, each of the following rules holds in finding derivatives.

**Sum Rule.** The derivative of the sum of f (x) and g (x) is the same as the sum of the derivative of f (x) and the derivative of g (x):

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

that is, for j (x) = f (x) + g (x), j’ (x) = f’ (x) + g’ (x)

**Difference Rule.** The derivative of the difference of f (x) and g (x) is the same as the difference of the derivative of f (x) and the derivative of g (x):

$\frac{d}{dx} [f (x) – g (x)] = \frac{d}{dx} [f (x)] – \frac{d}{dx} [g (x)]$ ;

that is, for j (x) = f (x) – g (x), j’ (x) = f’ (x) – g’ (x)

**Constant Multiple Rule.** The derivative of a constant k multiplied by f (x) is the same as the constant multiplied by the derivative of f (x):

$\frac{d}{dx}[k f(x) ]= k \frac{d}{dx}[ f(x)]$

that is, for j (x) =k f(x), j’ (x) =k f'(x)

**Proof**

We will only prove the sum rule because the rest can be proven in a similar manner.

Let f(x) and g(x) be differentiable functions. We set j(x)=f(x)+g(x). By the limit definition of derivatives, we will have

$j'(x)= = \frac{j(x+h)-j(x)}{h}$

By substitution of j(x+h)= f(x+h)+ g(x+h) and j(x)= f(x)+g(x), we will have the result

$j'(x)= \frac{[f(x+h)+ g(x+h)]-[f(x)+g(x)]}{h}$

By regrouping, we can obtain

$j'(x)= \frac{f (x+h)-f (x)}{h}+ \frac{g (x+h)-g (x)}{h}$

Applying the sum law definition of limits

$j'(x)= \frac{f (x+h)-f (x)}{h}+\lim_{h\rightarrow0} \frac{g (x+h)-g (x)}{h}$

$j'(x)= f'(x)+g'(x)$

**Example #1**

Let f(x)= 5x and g (x)=3x^{3}. Find the derivative of f(x)+g (x) and f(x)-g (x)

**Solution**

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

$\frac{d}{dx}[ f (x) + g (x)] = \frac{d}{dx} 5x + \frac{d}{dx} 3x^3$

$\frac{d}{dx}[ f (x) + g (x)] = 5+ 9x^2$

Therefore, the derivative of f(x)+g (x) is 5+ 9x^{2}.

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

$\frac{d}{dx}[ f (x) – g (x)] = \frac{d}{dx} 5x – \frac{d}{dx} 3x^3$

$\frac{d}{dx}[ f (x) – g (x)] = 5 – 9x^2$

Therefore, the derivative of f (x)+g (x) is 5 – 9x^{2}.

**Example #2**

Let f(x)=5(x). Find the derivative of 5f(x).

**Solution**

$\frac{d}{dx}[ k f(x)] = k \frac{d}{dx} [\:f(x)\:]$ Where k is the constant 5.

$\frac{d}{dx}[ k f(x)] = 5 \frac{d}{dx}5x$

$\frac{d}{dx}[ k f(x)] = 5 (5)$

$\frac{d}{dx}[ k f(x)] = 25$

Therefore, the derivative of 5f(x) is **25.**

**Product Rule**

The product rule says that the derivative of the product of two functions is the sum of the product of the first function and the derivative of the second function and the product of the second function and derivative of the first function.

Most of us may think that the derivative of the product of two functions is the product of the derivatives, similar to the sum and difference rules. But, the product rule does not work that way. For example, the derivative of f (x)=x^{2} is f’ (x) = 2x and is not $\frac{d}{dx} (x) ∙ \frac{d}{dx} (x)$ = 1 ∙ 1 = 1.

We can restate the product rule as follows.

Let f (x) and g (x) be differentiable functions.

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

That is, if j (x) = f (x) g(x), then j’ (x) = f (x) g’ (x) + g (x) f'(x).

**Proof**

Suppose f (x) and g (x) are both differentiable functions.

Let us apply the limit definition of the derivative to j (x) = f (x) g (x), to obtain

$j’ (x)= \frac{f (x + h) g (x + h) – f (x) g (x)}{h}$

The let us add and subtract f (x) g (x + h) in the numerator, so we can have

$j’ (x)= \frac{f(x+h) g (x+h)-f(x)g(x+h)+f(x) g(x+h) -f(x) g(x)}{h}$

Applying the sum law for limits, we can rewrite the derivative as

$j’ (x)= \frac{f(x+h) g (x+h)-f(x)g(x+h)}{h} +\lim_{h\rightarrow0} \frac{f(x) g(x+h) – f(x) g(x)}{h}$

Simplifying and rearranging the terms, the result would be

$j’ (x)= \frac{f(x+h) -f(x)}{h} +\lim_{h\rightarrow0} f(x) \frac{g(x+h) – g(x)}{h}$

By using the continuity of g(x), the definition of the derivatives of f(x) and g(x), and applying the limit laws, we will come up with the product rule.

**Example**

Find the derivative of f (x) = (2x + 1)( x + 2).

**Solution**

f’ (x) = (2x + 1) $\frac{d}{dx}$ (x + 2) + (x + 2) $\frac{d}{dx}$ (2x + 1)$

f’ (x) = (2x + 1)(1) + (x + 2) (2)

f’ (x) = 2x + 1 + 2x + 4

f’ (x) = 4x + 5

Therefore, the derivative of f(x)=(2x+1) (x+2) is f’ (x) = 4x + 5.

**Quotient Rule**

The derivative of the quotient of two functions is not the quotient of their derivatives. For example, the derivative of $\frac{d}{dx}$ x^{2}= 2x and is not $\frac{\frac{d}{dx} x^3}{\frac{d}{dx} x}=\frac{3x^2}{1}$=3 x^{2}. The quotient rule states that if a function is of the form $\frac{f(x)}{g(x)}$, then the derivative is the difference between the product of the denominator and the derivative of the numerator, and the product of the numerator and the derivative of the denominator, all over the square of the denominator.

We can restate the quotient rule as follows.

Let f (x) and g (x) be differentiable functions. Then,

$\frac{d}{dx} \left [ \frac{f (x)}{g (x)}\right ] = \frac{\frac{d}{dx} f (x) g (x) – \frac{d}{dx} g (x) f (x)}{g(x)^2}$

The proof of the quotient rule of derivatives can be shown in a similar manner as the proof of the product rule. So we will not proceed with it. Instead, we will use it in proving the extended power rule with negative exponents.

**Extended Power Rule**

The extended power rule states that if k is a negative exponent, then,

$\frac{d}{dx} (x^k) = kx^{k-1}$

**Proof**

Let n = -k, where k is a negative integer so that n is a positive integer with k =-n. We know that for every positive integer n, x^{-n} = $\frac{1}{x^n}$ with the application of the laws of exponents.

Now let us apply the quotient rule to find the derivative of x^{-n} by setting up f (x) = 1 and g (x) = x^{n} . In this case, f’ (x) = 0 and g’ (x) = nx^{n-1}. So,

$\frac{d}{dx} (x^{-n}) = \frac{0 (x^n) – 1 (nx^{n-1})}{(x^n)^2}$

By simplifying, the result would be

$\frac{d}{dx} (x^{-n}) = \frac{-nx^{n-1}}{x^2} = -nx^{(n-1)-2n} = – nx^{-n-1}$

Since k= -n, we can substitute it to the above equation to finalize our proof.

$\frac{d}{dx} (x^k) = kx^{k-1}$

**Example**

Find the derivative of f (x) = x^{-5}.

**Solution**

f'(x) = -5x^{-5-1}

f'(x) = -5x^{-6}

Therefore, the derivative of f (x) = x^{-5} is f'(x) = -5x^{-6}**.**

## Solved Problems Using Derivative Rules

Let us now use what we have learned about the fundamental derivative rules in solving problems involving differentiation.

**Problem #1**

Find the derivative of h (x)= $\frac{2x + 3}{3x – 1}$

**Solution**

Differentiation process | Step-by-step explanation |

h (x)= $\frac{2x + 3}{3x – 1}$ | Set up the function that you will differentiate. Since we will differentiate a quotient, we will use the quotient rule. |

h (x)= $\frac{[(2x + 3) \frac{d}{dx} (3x – 1)] – [(3x – 1) \frac{d}{dx} (2x + 3)}{(3x – 1)^2}$ | Write the derivative applying the quotient rule. |

h (x)= $\frac{[(2x + 3) (3)] – [(3x – 1) (2)]}{(3x – 1)^2}$ | Simplify the numerator by finding the required derivatives. Using the sum and difference rule,ddx 3x – 1=3 andddx 2x + 3=2 |

h’ (x)= $\frac{(6x + 9) – (6x – 2)}{9x^2-6x+1}$ | Simplify the denominator. |

h’ (x)= $\frac{11}{9x^2-6x+1}$ | Simplify further the numerator by combining like terms. |

Therefore, derivative of h (x)= $\frac{2x + 3}{3x – 1}$ is h’ (x)= $\frac{11}{9x^2-6x+1}$ |

**Problem #2**

Given f(x) = (3x^{2} – 1)(x^{2} + x +2), find the derivative of f(x).

**Solution**

Differentiation process | Step-by-step explanation |

f(x) = (3x^{2} – 1)(x^{2} + x +2) | Set up the function that you will differentiate. Since we will differentiate a product, we will apply the product rule. |

f'(x)= (3x^{2} – 1)$\frac{d}{dx}$ (x^{2} + x +2)+(x^{2} + x +2)$\frac{d}{dx}$(3x^{2} – 1) | Write the derivative applying the product rule. |

f'(x)= (3x^{2} – 1)(2x + 1) + (x^{2} + x +2) (6x) | Simplify the numerator by finding the required derivatives. Using the sum and difference rule, $\frac{d}{dx}$ (x^{2} + x +2) = 2x + 1 and $\frac{d}{dx}$ (3x^{2} – 1) = 6x. |

f'(x)= (6x^{3} + 3x^{2} – 2x – 1) + (6x^{3}+6x^{2}+ 12x) | Simplify each term by multiplying. |

f'(x)= 12x^{3} + 9x^{2} + 10x – 1 | Simplify further by combining like terms. |

**Problem #3**

The sales (in millions of dollars) of a song by an international artist n years from the date of its release is given by g(n)= $\frac{4n}{n^2+1}$. Find the rate of change (g'(n)) of the sales at time n and determine how fast the rate is changing two years from the date of release.

**Solution**

The rate of change of the sales at time n is given by g’^{n}.

Differentiation process | Step-by-step explanation |

g'(n) = $\frac{d}{dn} (\frac{4n}{n^2 + 1})$ | Set up the differentiation equation. |

g'(n) = $4\frac{d}{dn} (\frac{n}{n^2 + 1})$ | By the constant multiple rule, $\frac{d}{dn} (\frac{4n}{n^2 + 1}) = 4\frac{d}{dn} (\frac{n}{n^2 + 1})$ |

g'(n) = $4 \left (\frac{(n^2 + 1)\frac{d}{dn} n – n \frac{d}{dn}(n^2 + 1)}{(n^2 + 1)^2} \right )$ | Apply the quotient rule to solve for $\frac{d}{dn} (\frac{n}{n^2 + 1})$ |

g'(n) = $4 \left ( \frac{(n^2 + 1)1-n2n}{(n^2 + 1)^2} \right )$ | Simplify the numerator by solving for the required derivatives. $\frac{d}{dn}$n=1 and $\frac{d}{dn}$(n^{2} + 1)=2n |

g'(n) =$\frac{4(1-n^2)}{(n^2 + 1)^2}$ | Simplify the numerator. |

Thus, the rate of change of the sales at time n is g'(n) =$\frac{4(1-n^2)}{(n^2 + 1)^2}$ | |

g'(3) =$ \frac{4(1-2^2)}{(2^2 + 1)^2}$. | To determine how fast is the rate changing a year from the date of release, we substitute n=2 to the equation |

g'(1) =$\frac{-12}{25}$= -0.48 | Evaluate the equation. |

Therefore, the sales are decreasing at the rate of $480 000 two years from the date of release. |

## What is the application of derivatives in real life?

The derivative is the rate of change of a function with respect to another quantity. Some of its applications are checking whether a function is increasing or decreasing, determining the tangent/normal equation, determining the maximum and minimum values from a graph, determining displacement-motion problems, determining velocity given displacement, and determining acceleration given a displacement.

Derivatives have been used on both a small and large scale. The concept of derivatives is used in several ways, such as temperature change or the rate of change of measurements of an object depending on the conditions, and so on. Consider speed which is defined as the instantaneous rate of change in the distance traveled by an object at a given time. The velocity of an object is the first derivative of its displacement. The acceleration of an object is represented by the second derivative of displacement. The object’s jerk is the third derivative of displacement.

Another example is the rate of change of the volume of a spherical balloon with respect to its decreasing radius. We can represent the rate of change as dy/dx, where dy represents the rate of change of volume of the balloon and dx represents the change of radius of the balloon.

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