Lesson Explainer: Power Rule of Derivatives Mathematics • Higher Education

In this explainer, we will learn how to use the power rule of derivatives and the derivative of a sum of functions to find the derivatives of polynomials and general power functions.

We begin by recalling the definition of the derivative.

Definition: Derivative of a Function

The derivative of a function 𝑓 is defined as 𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„ŽοŽ˜ο‚β†’οŠ¦lim at the points where the limit exists.

Using such a definition to calculate derivatives can be rather tedious. We would therefore like to find some rules which we can use to simplify finding the derivatives of functions. In this explainer, we will consider some of the key rules which will enable us to differentiate a whole collection of functions.

We will begin by considering one of the simplest categories of functions: constant functions.

Example 1: Derivative of a Constant

Find dd𝑦π‘₯, given that 𝑦=βˆ’60.

Answer

Recall the definition of the derivative for a general function 𝑦=𝑓(π‘₯): ddlim𝑦π‘₯=𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Ž.οŽ˜ο‚β†’οŠ¦

Substituting in 𝑓(π‘₯)=βˆ’60, we have ddlimlim𝑦π‘₯=βˆ’60βˆ’(βˆ’60)β„Ž=0=0.ο‚β†’οŠ¦ο‚β†’οŠ¦

Therefore, the derivative of 𝑦=βˆ’60 is zero for all values of π‘₯.

The last example demonstrated a general rule of derivatives that the derivative of a constant is zero.

Rule: Constant Rule

For a constant 𝑐, ddπ‘₯(𝑐)=0.

We can now consider functions of the form 𝑓(π‘₯)=π‘₯.

Example 2: The Power Rule for Positive Integers

  1. Find the derivative of 𝑓(π‘₯)=π‘₯ by first principles.
  2. Find the derivative of 𝑓(π‘₯)=π‘₯ by first principles.
  3. Find the derivative of 𝑓(π‘₯)=π‘₯ by first principles.
  4. By considering the pattern, what is the derivative of 𝑓(π‘₯)=π‘₯?

Answer

Part 1

Recall the definition of the derivative for a general function 𝑦=𝑓(π‘₯): 𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Ž.οŽ˜ο‚β†’οŠ¦lim

Using the function 𝑓(π‘₯)=π‘₯, we have 𝑓(π‘₯)=π‘₯+β„Žβˆ’π‘₯β„Ž=β„Žβ„Ž.οŽ˜οŠ§ο‚β†’οŠ¦ο‚β†’οŠ¦limlim

Since β„Žβ‰ 0, we can cancel the common factor of β„Ž from the numerator and denominator to get 𝑓(π‘₯)=1=1.οŽ˜οŠ§ο‚β†’οŠ¦lim

Part 2

Using the function 𝑓(π‘₯)=π‘₯ and the definition of the derivative, we have 𝑓(π‘₯)=(π‘₯+β„Ž)βˆ’π‘₯β„Ž.οŽ˜οŠ¨ο‚β†’οŠ¦οŠ¨οŠ¨lim

Expanding the parentheses in the numerator, we have 𝑓(π‘₯)=π‘₯+2β„Žπ‘₯+β„Žβˆ’π‘₯β„Ž=2β„Žπ‘₯+β„Žβ„Ž.οŽ˜οŠ¨ο‚β†’οŠ¦οŠ¨οŠ¨οŠ¨ο‚β†’οŠ¦οŠ¨limlim

Since β„Žβ‰ 0, we can cancel the common factor of β„Ž from the numerator and denominator to get 𝑓(π‘₯)=(2π‘₯+β„Ž)=2π‘₯.οŽ˜οŠ¨ο‚β†’οŠ¦lim

Part 3

Using the function 𝑓(π‘₯)=π‘₯ and the definition of the derivative, we have 𝑓(π‘₯)=(π‘₯+β„Ž)βˆ’π‘₯β„Ž.οŽ˜οŠ©ο‚β†’οŠ¦οŠ©οŠ©lim

Expanding the parentheses in the numerator, we have 𝑓(π‘₯)=π‘₯+3β„Žπ‘₯+3β„Žπ‘₯+β„Žβˆ’π‘₯β„Ž=3β„Žπ‘₯+3β„Žπ‘₯+β„Žβ„Ž.οŽ˜οŠ©ο‚β†’οŠ¦οŠ©οŠ¨οŠ¨οŠ©οŠ©ο‚β†’οŠ¦οŠ¨οŠ¨οŠ©limlim

Since β„Žβ‰ 0, we can cancel the common factor of β„Ž from the numerator and denominator to get 𝑓(π‘₯)=ο€Ή3π‘₯+3β„Žπ‘₯+β„Žο…=3π‘₯.οŽ˜οŠ©ο‚β†’οŠ¦οŠ¨οŠ¨οŠ¨lim

Part 4

Putting together the answers from the first three parts of the question, we have ddddddπ‘₯(π‘₯)=1,π‘₯ο€Ήπ‘₯=2π‘₯,π‘₯ο€Ήπ‘₯=3π‘₯.

We can see that when we take the derivative of a power of π‘₯, the power reduces by one. In addition to this, we find that there is a constant equal to the original power. Hence, ddπ‘₯(π‘₯)=𝑛π‘₯.

The previous example leads us to a general rule for differentiating powers of π‘₯; we call it the power rule.

Rule: Power Rule for Positive Integers

For any positive integer 𝑛, ddπ‘₯(π‘₯)=𝑛π‘₯.

To prove the power rule for positive integers, we will need to use the binomial theorem. Recall that the binomial theorem enables us to expand binomials raised to any positive integer power in particular: (π‘Ž+𝑏)=π‘Ž+𝑛1ο‹π‘Žπ‘+𝑛2ο‹π‘Žπ‘+β‹―+ο€Ώπ‘›π‘›βˆ’1ο‹π‘Žπ‘+𝑏.

Let 𝑓(π‘₯)=π‘₯; using the definition of the derivative, we have 𝑓(π‘₯)=(π‘₯+β„Ž)βˆ’π‘₯β„Ž.οŽ˜ο‚β†’οŠ¦οŠοŠlim

Using the binomial theorem, we can expand (π‘₯+β„Ž) as follows: 𝑓(π‘₯)=ο€Ίπ‘₯+π‘₯β„Ž+π‘₯β„Ž+β‹―+π‘₯β„Ž+β„Žο†βˆ’π‘₯β„Ž=π‘₯β„Ž+π‘₯β„Ž+β‹―+π‘₯β„Ž+β„Žβ„Ž.οŽ˜ο‚β†’οŠ¦οŠοŠοŠ§οŠοŠ±οŠ§οŠοŠ¨οŠοŠ±οŠ¨οŠ¨οŠοŠοŠ±οŠ§οŠ¨οŠοŠ±οŠ§οŠοŠο‚β†’οŠ¦οŠοŠ§οŠοŠ±οŠ§οŠοŠ¨οŠοŠ±οŠ¨οŠ¨οŠοŠοŠ±οŠ§οŠ¨οŠοŠ±οŠ§οŠlimlim

Since β„Žβ‰ 0, we can cancel the common factor of β„Ž from the numerator and denominator to get 𝑓(π‘₯)=𝑛1π‘₯+𝑛2π‘₯β„Ž+β‹―+ο€Ώπ‘›π‘›βˆ’1π‘₯β„Ž+β„ŽοŠ.οŽ˜ο‚β†’οŠ¦οŠοŠ±οŠ§οŠοŠ±οŠ¨οŠ¨οŠοŠ±οŠ¨οŠοŠ±οŠ§lim

Taking the limit as β„Žβ†’0, the only term without a positive power of β„Ž in it is 𝑛1π‘₯. Hence, 𝑓(π‘₯)=𝑛1π‘₯=𝑛π‘₯.

Notice that if we set 𝑛=0, we have a constant function and the power rule tells us that the derivative is zero in agreement with our initial rule regarding the derivatives of constant functions.

Although this is a powerful rule that enables us to consider derivatives of a much larger class of functions, it is still rather limiting since we are not able to use this to differentiate functions made of multiple terms such as polynomials. We would, therefore, like to know rules about how to differentiate functions multiplied by constants and functions formed as the sum of two simpler functions. We will begin with considering the derivative of a function 𝑓(π‘₯)=𝑐𝑔(π‘₯), where 𝑐 is a constant and 𝑔 is a differentiable function.

Using the definition of the derivative, we have 𝑓(π‘₯)=𝑐𝑔(π‘₯+β„Ž)βˆ’π‘π‘”(π‘₯)β„Ž.οŽ˜ο‚β†’οŠ¦lim

Factoring out the common factor of 𝑐, we have 𝑓(π‘₯)=𝑐𝑔(π‘₯+β„Ž)βˆ’π‘”(π‘₯)β„Žο‰.οŽ˜ο‚β†’οŠ¦lim

Given that 𝑔 is differentiable, we can use the rules of finite limits to rewrite this as 𝑓(π‘₯)=𝑐𝑔(π‘₯+β„Ž)βˆ’π‘”(π‘₯)β„Žο‰=𝑐𝑔(π‘₯).οŽ˜ο‚β†’οŠ¦οŽ˜lim

Rule: Constant Multiple Rule

For a constant 𝑐, ddddπ‘₯(𝑐𝑓(π‘₯))=𝑐π‘₯𝑓(π‘₯).

We will now consider the derivative of a function defined as the sum of two differentiable functions. Let 𝐹(π‘₯)=𝑓(π‘₯)+𝑔(π‘₯); then, using the definition of the derivative, we have 𝐹(π‘₯)=𝐹(π‘₯+β„Ž)βˆ’πΉ(π‘₯)β„Ž=𝑓(π‘₯+β„Ž)+𝑔(π‘₯+β„Ž)βˆ’(𝑓(π‘₯)+𝑔(π‘₯))β„Ž=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Ž+𝑔(π‘₯+β„Ž)βˆ’π‘”(π‘₯)β„Žο‰.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦limlimlim

Since both 𝑓 and 𝑔 are differentiable, we have 𝐹(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„Ž+𝑔(π‘₯+β„Ž)βˆ’π‘”(π‘₯)β„Ž=𝑓(π‘₯)+𝑔(π‘₯).οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦οŽ˜οŽ˜limlim

Using the rule in conjunction with the constant multiple rule, we can see ddddπ‘₯(𝑓(π‘₯)βˆ’π‘”(π‘₯))=π‘₯(𝑓(π‘₯)+(βˆ’1)𝑔(π‘₯)).

Using the sum rule, we have ddddddπ‘₯(𝑓(π‘₯)βˆ’π‘”(π‘₯))=π‘₯𝑓(π‘₯)+π‘₯(βˆ’1)𝑔(π‘₯).

We can now use the constant multiple rule to get ddddddddddπ‘₯(𝑓(π‘₯)βˆ’π‘”(π‘₯))=π‘₯𝑓(π‘₯)+(βˆ’1)π‘₯𝑔(π‘₯)=π‘₯𝑓(π‘₯)βˆ’π‘₯𝑔(π‘₯).

We summarize these results below.

Rule: Sum and Difference Rule

For two differentiable functions 𝑓 and 𝑔, ddddddπ‘₯(𝑓(π‘₯)±𝑔(π‘₯))=π‘₯𝑓(π‘₯)Β±π‘₯𝑔(π‘₯).

We now know the necessary tools to be able to differentiate polynomials.

Example 3: Derivatives of Polynomials

Given that 𝑓(π‘₯)=βˆ’π‘₯+π‘šπ‘₯+1, determine π‘š if 𝑓(3)=1.

Answer

Since we know the value of the derivative at a certain point, we should begin by finding an expression for the derivative. Using the sum and difference rule, we have 𝑓(π‘₯)=π‘₯ο€Ήβˆ’π‘₯+π‘₯(π‘šπ‘₯)+π‘₯(1).dddddd

We can now use the constant multiple rule to get 𝑓(π‘₯)=βˆ’π‘₯ο€Ήπ‘₯+π‘šπ‘₯(π‘₯)+π‘₯(1).dddddd

We can now apply the power rule, ddπ‘₯(π‘₯)=𝑛π‘₯, to each term as follows: 𝑓(π‘₯)=βˆ’2π‘₯+π‘š.

Since 𝑓(3)=1, we have 1=βˆ’2(3)+π‘š=βˆ’6+π‘š.

Therefore, π‘š=7.

The last example demonstrated how to carefully apply the sum, difference, and constant multiple rules. In general, we will not need to give the full details of applying these rules when differentiating functions. In later examples, we will not use the same level of detail when applying these rules.

In the next example, we will consider whether the power rule generalizes to include negative powers.

Example 4: Differentiating Negative Powers

Find the derivative of 𝑓(π‘₯)=π‘₯.

Answer

Using the definition of the derivative 𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„ŽοŽ˜ο‚β†’οŠ¦lim with 𝑓(π‘₯)=π‘₯, we have 𝑓(π‘₯)=1β„Žο€Ό1π‘₯+β„Žβˆ’1π‘₯.οŽ˜ο‚β†’οŠ¦lim

We can express this as a single fraction as follows: 𝑓(π‘₯)=1β„Žο€½π‘₯βˆ’(π‘₯+β„Ž)π‘₯(π‘₯+β„Ž)=1β„Žο€½βˆ’β„Žπ‘₯+π‘₯β„Žο‰.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦οŠ¨limlim

Since β„Žβ‰ 0, we can cancel it from the numerator and denominator as follows: 𝑓(π‘₯)=ο€Όβˆ’1π‘₯+π‘₯β„Žοˆ.οŽ˜ο‚β†’οŠ¦οŠ¨lim

Using the laws of finite limits, we have 𝑓(π‘₯)=βˆ’1ο€Ήπ‘₯+π‘₯β„Žο…=βˆ’1π‘₯+π‘₯β„Ž.οŽ˜ο‚β†’οŠ¦οŠ¨οŠ¨ο‚β†’οŠ¦limlim

Taking the limit as β„Žβ†’0, we have 𝑓(π‘₯)=βˆ’1π‘₯.

From the previous example, we saw that ddπ‘₯ο€Ήπ‘₯=βˆ’1π‘₯.

This can also be written as ddπ‘₯ο€Ήπ‘₯=(βˆ’1)π‘₯ which is the same form as the power rule if we set 𝑛=βˆ’1. This is not an accident; and in fact, the power rule generalizes to all integers.

Rule: Power Rule for Integer Powers

For any integer 𝑛, ddπ‘₯(π‘₯)=𝑛π‘₯.

We will not prove this version of the power rule right now. However, it is straightforward to prove once we have been introduced to more advanced rules of differentiation. In the next example, we will consider whether the power rule generalizes further to include fractional exponents.

Example 5: Differentiating Fractional Powers

Find the derivative of 𝑓(π‘₯)=√π‘₯.

Answer

Using the definition of the derivative 𝑓(π‘₯)=𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯)β„ŽοŽ˜ο‚β†’οŠ¦lim with 𝑓(π‘₯)=π‘₯, we have 𝑓(π‘₯)=√π‘₯+β„Žβˆ’βˆšπ‘₯β„Ž.οŽ˜ο‚β†’οŠ¦lim

Since this limit is tending to 00, we can use the algebraic technique of multiplying the numerator and denominator by the conjugate of the numerator to manipulate this fraction into a form where we can evaluate the limit. Hence, 𝑓(π‘₯)=ο€βˆšπ‘₯+β„Žβˆ’βˆšπ‘₯β„ŽΓ—βˆšπ‘₯+β„Ž+√π‘₯√π‘₯+β„Ž+√π‘₯=π‘₯+β„Žβˆ’π‘₯β„Žο€»βˆšπ‘₯+β„Ž+√π‘₯=β„Žβ„Žο€»βˆšπ‘₯+β„Ž+√π‘₯.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦ο‚β†’οŠ¦limlimlim

Since β„Žβ‰ 0, we can cancel it from the numerator and denominator as follows: 𝑓(π‘₯)=1√π‘₯+β„Ž+√π‘₯.οŽ˜ο‚β†’οŠ¦lim

We can now apply the rules of finite limits to get 𝑓(π‘₯)=1ο€»βˆšπ‘₯+β„Ž+√π‘₯=1√π‘₯+β„Ž+√π‘₯.οŽ˜ο‚β†’οŠ¦ο‚β†’οŠ¦limlim

Since limο‚β†’οŠ¦βˆšπ‘₯+β„Ž=√π‘₯, we have 𝑓(π‘₯)=12√π‘₯.

The previous example showed us that ddπ‘₯√π‘₯=12√π‘₯.

We can rewrite this using index notation as ddπ‘₯ο€½π‘₯=12π‘₯, which is exactly the formula we would have gotten were we to set 𝑛=12 in the power rule. This is once again no accident. In fact, not only does the power rule generalize to rational powers, but it also generalizes to any real power.

Rule: Power Rule for General Powers

For any real number π‘Ÿ, ddπ‘₯(π‘₯)=π‘Ÿπ‘₯.

We will not prove this most general form of the power rule here. The general version of power rule will be simple to prove once we have introduced the logarithmic differentiation technique.

In the last couple of examples, we will apply the power rule and the sum, difference, and constant multiple rules to find derivatives of more general functions.

Example 6: Using the Power Rule to Differentiate

Given that 𝑦=15βˆ’1π‘₯+13π‘₯, find π‘¦οŽ˜.

Answer

Using properties of derivatives, we can differentiate each term independently as follows: 𝑦=π‘₯(15)βˆ’π‘₯ο€Ό1π‘₯+13π‘₯ο€Ήπ‘₯.dddddd

We can now apply the power rule, ddπ‘₯(π‘₯)=𝑛π‘₯, to each term as follows: 𝑦=0βˆ’(βˆ’6)π‘₯+13(20)π‘₯=6π‘₯+203π‘₯.

Example 7: Using the Power Rule to Differentiate

Given that 𝑦=3π‘₯+4π‘₯+π‘₯βˆ’2√π‘₯οŠͺ, find π‘¦οŽ˜.

Answer

Before we attempt to differentiate this function, we should simplify the expression. Since the denominator consists of a single term, we can use the rules of powers to simplify the expression as follows: 𝑦=3π‘₯π‘₯+4π‘₯π‘₯+π‘₯π‘₯βˆ’2π‘₯=3π‘₯+4π‘₯+π‘₯βˆ’2π‘₯.οŠͺ

We can now use the power rule, ddπ‘₯(π‘₯)=𝑛π‘₯, to differentiate each term as follows: 𝑦=3ο€Ό112π‘₯+4ο€Ό72π‘₯+ο€Ό32π‘₯βˆ’2ο€Όβˆ’12π‘₯=332π‘₯+14π‘₯+32π‘₯+π‘₯.

Let's recap a few important concepts from this explainer.

Key Points

  • The power rule for differentiation states that, for any real number π‘Ÿ, ddπ‘₯(π‘₯)=π‘Ÿπ‘₯.
  • The sum and difference rule states that, for two differentiable functions 𝑓 and 𝑔, ddddddπ‘₯(𝑓(π‘₯)±𝑔(π‘₯))=π‘₯𝑓(π‘₯)Β±π‘₯𝑔(π‘₯).
  • The constant multiple rule states that, for a constant 𝑐, ddddπ‘₯(𝑐𝑓(π‘₯))=𝑐π‘₯𝑓(π‘₯).
  • Using these rules of differentiation, it is possible to find the derivatives of a wide class of functions.

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