Lesson Explainer: Second- and Higher-Order Derivatives Mathematics • Higher Education

In this explainer, we will learn how to find second- and higher-order derivatives of a function including using differentiation rules.

The derivative of a function tells us information about the rate of change of the function. This is an incredibly useful concept and it is used in a lot of real-world problems such as population growth and optimization. However, we do not need to stop there, we can also ask the question, “how is the rate of change itself changing?” We can determine this by differentiating the first derivative; we call this the second derivative. In fact, we can keep differentiating to find higher-order derivatives.

These higher-order derivatives are also useful in many real-world problems including spring–mass systems and the two-body problem. There are two main notations for derivatives: Leibniz’s notation and prime notation (sometimes referred to as Lagrange’s notation). Let’s start by extending these notations to higher-order derivatives.

Definition: Higher-Order Derivatives

As long as each derivative exists, we can differentiate functions any number of times and we can represent the nth derivative of a function 𝑦=𝑓(𝑥) as dd𝑦𝑥 or in two ways using prime notation as 𝑓(𝑥)𝑓(𝑥),()or where the prime symbol appears 𝑛 times.

We can evaluate the higher-order derivatives of a function at a value of 𝑥. For example, the second derivative evaluated at 𝑥=1 could be written as 𝑦𝑥𝑓(1).ddor

In our first example, we will find the second derivative of a polynomial function.

Example 1: Finding the Second Derivative of a Polynomial Function

Given that 𝑦=6𝑥+3𝑥7𝑥+6, determine dd𝑦𝑥.

Answer

dd𝑦𝑥 means we need to differentiate 𝑦 twice with respect to 𝑥. 𝑦 is a polynomial in 𝑥, and we can differentiate polynomials term by term using the power rule for differentiation: for 𝑎, 𝑛, dd𝑥(𝑎𝑥)=𝑎𝑛𝑥.

We can start by differentiating once to find dd𝑦𝑥: dddddddddddddd𝑦𝑥=𝑥6𝑥+3𝑥7𝑥+6=𝑥6𝑥+3𝑥7𝑥+6𝑥=𝑥6𝑥+𝑥3𝑥+𝑥7𝑥+𝑥6𝑥=6(5)𝑥+3(2)𝑥+(7)(1)𝑥+6(0)𝑥=30𝑥+6𝑥7.

Next, to find dd𝑦𝑥, we need to differentiate this expression again: dddddddddddddd𝑦𝑥=𝑥𝑦𝑥=𝑥30𝑥+6𝑥7=𝑥30𝑥+𝑥6𝑥+𝑥7𝑥=30(4)𝑥+6(1)𝑥+(7)(0)𝑥=120𝑥+6=620𝑥+1.

Hence, dd𝑦𝑥=620𝑥+1.

In our second example, we determine the value of the second derivative of a function at a given point.

Example 2: Finding the Value of the Second Derivative of a Function at a Given Point

Determine the value of the second derivative of the function 𝑦=12𝑥8𝑥 at (1,4).

Answer

We find the second derivative of 𝑦 by differentiating function 𝑦 to obtain dd𝑦𝑥 and then differentiating again to obtain dd𝑦𝑥. To do this, we will start by rewriting our expression: 𝑦=12𝑥8𝑥=12𝑥8𝑥.

We can then differentiate this expression by using the power rule for differentiation: for 𝑘,𝑛, dd𝑥(𝑘𝑥)=𝑘𝑛𝑥.

Hence, dddddddd𝑦𝑥=𝑥12𝑥8𝑥=𝑥12𝑥𝑥8𝑥=12(1)𝑥8(1)𝑥=12+8𝑥.

We can differentiate again to find dd𝑦𝑥: dddddddddddd𝑦𝑥=𝑥𝑦𝑥=𝑥12+8𝑥=𝑥12𝑥+𝑥8𝑥=12(0)𝑥+8(2)𝑥=16𝑥.

Finally, the question wants us to find the value of the second derivative at the point (1,4); this is when 𝑥=1. We substitute this value into the expression: 𝑦𝑥=16(1)=16.dd

Hence, the value of the second derivative of the function at the point (1,4) is 16.

In our next example, we will find the second derivative of a rational function.

Example 3: Finding the Second Derivative of Rational Functions

Given that 𝑦=3𝑥52𝑥+7, determine dd𝑦𝑥.

Answer

To find dd𝑦𝑥, we need to differentiate 𝑦 with respect to 𝑥 twice. Since we are given a rational function, we can find the first derivative by using the quotient rule.

If 𝑢 and 𝑣 are differentiable functions, then dd𝑥𝑢𝑣=𝑣𝑢𝑣.dddd

We have 𝑢=3𝑥5,𝑣=2𝑥+7.

Since these are both polynomials, both functions are differentiable and, hence, we can differentiate them by using the power rule for differentiation: for 𝑎,𝑛, dd𝑥(𝑎𝑥)=𝑎𝑛𝑥.

This gives dddddddd𝑢𝑥=𝑥3𝑥5=6𝑥,𝑣𝑥=𝑥2𝑥+7=4𝑥.

Substituting our expressions for 𝑢, 𝑣, dd𝑢𝑥, and dd𝑣𝑥 into the quotient rule gives dd𝑦𝑥=2𝑥+7(6𝑥)3𝑥5(4𝑥)(2𝑥+7)=12𝑥+42𝑥12𝑥+20𝑥(2𝑥+7)=62𝑥(2𝑥+7).

We need to differentiate this expression to find dd𝑦𝑥. Since this is a rational function, we will do this once again by using the quotient rule. This gives 𝑢=62𝑥,𝑣=2𝑥+7.

To help us differentiate 𝑣, we will distribute the exponent: 𝑣=2𝑥+7=4𝑥+28𝑥+49.

We can then differentiate both functions by using the power rule for differentiation: dddddddd𝑢𝑥=𝑥(62𝑥)=62𝑣𝑥=𝑥4𝑥+28𝑥+49=16𝑥+56𝑥=8𝑥2𝑥+7.

Substituting our expressions for 𝑢, 𝑣, dd𝑢𝑥, and dd𝑣𝑥 into the quotient rule gives dd𝑦𝑥=2𝑥+7(62)(62𝑥)8𝑥2𝑥+7(2𝑥+7)=622𝑥+72𝑥+78𝑥(2𝑥+7)=6276𝑥(2𝑥+7).

Hence, dd𝑦𝑥=6276𝑥(2𝑥+7).

In our next example, we will use knowledge of the higher-order derivatives of a function to determine the values of coefficients in the function.

Example 4: Finding the Unknown Coefficients in the Expression of a Function given the Values of the Second and Third Derivatives of the Function

Given that 𝑦=𝑎𝑥+𝑏𝑥, 𝑦=18, and 𝑦𝑥=14dd, find 𝑎 and 𝑏.

Answer

We want to determine the coefficients of both terms in our function 𝑦. We are given information about the third derivative of 𝑦, 𝑦, and the second derivative, dd𝑦𝑥.

Let’s start by finding expressions for these higher-order derivatives. Remember, to find dd𝑦𝑥, we need to differentiate 𝑦 twice, and to find 𝑦, we need to differentiate 𝑦 three times.

We can differentiate a polynomial by using the power rule for differentiation: for 𝑎,𝑛, dd𝑥(𝑎𝑥)=𝑎𝑛𝑥.

Applying this, we have dddd𝑦𝑥=𝑥𝑎𝑥+𝑏𝑥=3𝑎𝑥+2𝑏𝑥.

Differentiating again, we have dddd𝑦𝑥=𝑥3𝑎𝑥+2𝑏𝑥=6𝑎𝑥+2𝑏.

Then, differentiating one final time, we get dddd𝑦𝑥=𝑥(6𝑎𝑥+2𝑏)=6𝑎.

Remember, 𝑦=𝑦𝑥dd, and we know this is equal to 18.This gives 6𝑎=18𝑎=3.

We are also told 𝑦𝑥=14dd; this means when we evaluate the second derivative at 𝑥=2, we should get 14. Hence, 14=𝑦𝑥=6𝑎(2)+2𝑏=12𝑎+2𝑏.dd

Substituting 𝑎=3 into this equation gives 14=12(3)+2𝑏14=36+2𝑏22=2𝑏11=𝑏.

Therefore, we have shown 𝑎=3 and 𝑏=11.

In our next example, we will discover a pattern to determine a higher-order derivative of a trigonometric function.

Example 5: Finding the 𝑛th Derivative of an Expression

Find ddsin𝑥(𝑥) by finding the first few derivatives and observing the pattern that occurs.

Answer

We want to find ddsin𝑥(𝑥); this is the fifty-first derivative of sin𝑥 with respect to 𝑥. We could find this expression by differentiating this function fifty-one times; however, we will see there is a pattern in the derivatives.

Let’s start by recalling the following derivative results for trigonometric functions: ddsincosandddcossin𝑥(𝑥)=𝑥𝑥(𝑥)=𝑥.

So, this gives us the first derivative of sin𝑥: ddsincos𝑥(𝑥)=𝑥.

We find the second derivative of sin𝑥 by differentiating the first derivative with respect to 𝑥: ddsinddcossin𝑥(𝑥)=𝑥(𝑥)=𝑥.

Following the same method, we find that ddsinddsincosddsinddcossin𝑥(𝑥)=𝑥(𝑥)=𝑥,(𝑥(𝑥)=𝑥(𝑥)=𝑥.

Hence, the fourth derivative of the sine function is also the sine function. This means every four times we differentiate the sine function, we will get the sine function. Therefore, since 48 is a multiple of 4, ddsinddddsinsin𝑥(𝑥)=𝑥𝑥(𝑥)=𝑥.

To find the fifty-first derivative, we need to differentiate this three further times: ddsinddddsinddsincos𝑥(𝑥)=𝑥𝑥(𝑥)=𝑥(𝑥)=𝑥.

Hence, ddsincos𝑥(𝑥)=𝑥.

We may need to use derivative rules multiple times in order to find some higher-order derivatives as we will see in our next two examples.

Example 6: Finding the Third Derivative of Trigonometric Functions Using the Product Rule

Find the third derivative of the function 𝑦=44𝑥2𝑥sin.

Answer

To find the third derivative of a function, we need to differentiate it three times. Since we are given a product of two differentiable functions, we will find the first derivative by using the product rule:

If 𝑢(𝑥) and 𝑣(𝑥) are differentiable, then dd𝑥(𝑢(𝑥)𝑣(𝑥))=𝑢(𝑥)𝑣(𝑥)+𝑢(𝑥)𝑣(𝑥).

We will set 𝑢(𝑥)=44𝑥𝑣(𝑥)=2𝑥.andsin

We can differentiate each of these functions separately: 𝑢(𝑥)=44𝑣(𝑥)=22𝑥.andcos

Substituting these expressions into the product rule gives us ddsinsincossincos𝑥(44𝑥2𝑥)=442𝑥+(44𝑥)(22𝑥)=442𝑥+88𝑥2𝑥.

We need to differentiate this expression again to find the second derivative: ddddsincosddsinddcoscosddcos𝑦𝑥=𝑥(442𝑥+88𝑥2𝑥)=𝑥(442𝑥)+𝑥(88𝑥2𝑥)=882𝑥+𝑥(88𝑥2𝑥).

We will need to use the product rule for differentiation again; this time 𝑢(𝑥)=88𝑥𝑣(𝑥)=2𝑥.andcos

We can differentiate each function separately: 𝑢(𝑥)=88𝑣(𝑥)=22𝑥.andsin

Substituting these expressions into the product rule gives us ddcoscossincossin𝑥(88𝑥2𝑥)=882𝑥+88𝑥(22𝑥)=882𝑥176𝑥2𝑥.

Hence, ddcoscossincossin𝑦𝑥=882𝑥+882𝑥176𝑥2𝑥=1762𝑥176𝑥2𝑥.

We need to differentiate this once more to find the third derivative: ddddcossinddcosddsinsinddsin𝑦𝑥=𝑥(1762𝑥176𝑥2𝑥)=𝑥(1762𝑥)𝑥(176𝑥2𝑥)=3522𝑥𝑥(176𝑥2𝑥).

Since we already know ddsin𝑥(44𝑥2𝑥), we will rewrite this derivative: ddsinddsinsinddsinsinsincoscossin𝑦𝑥=3522𝑥𝑥(176𝑥2𝑥)=3522𝑥4𝑥(44𝑥2𝑥)=3522𝑥4(442𝑥+88𝑥2𝑥)=352𝑥2𝑥5282𝑥.

Therefore, ddcossin𝑦𝑥=352𝑥2𝑥5282𝑥.

Example 7: Finding the Second Derivative of Root Functions

Given 𝑦=𝑥9, find dd𝑦𝑥.

Answer

To find dd𝑦𝑥, we need to differentiate 𝑦 with respect to 𝑥 twice. Since we are given a composition of differentiable functions, we could differentiate this using the chain rule; however, we will use the general power rule.

If 𝑢(𝑥)is a differentiable function and 𝑘, then dd𝑥[𝑢(𝑥)]=𝑘𝑢(𝑥)[𝑢(𝑥)].

First, we rewrite our function: 𝑦=𝑥9=(𝑥9).

We set 𝑘=12 and 𝑢(𝑥)=𝑥9; since 𝑢(𝑥) is linear, 𝑢(𝑥) will be the slope of this line. Hence, 𝑢(𝑥)=1.

Substituting these expressions into the general power rule gives us dd𝑦𝑥=12(1)(𝑥9)=12(𝑥9).

To find the second derivative, we need to differentiate this expression: dddddd𝑦𝑥=𝑥12(𝑥9)=12𝑥(𝑥9).

We can do this by using the general power rule; we set 𝑢(𝑥)=𝑥9 and 𝑘=12, giving us dddd𝑦𝑥=12𝑥(𝑥9)=1212(1)(𝑥9)=14(𝑥9).

In our final example, we will use the chain rule to find the second derivative of a trigonometric function.

Example 8: Finding the Second Derivative of Trigonometric Functions Using the Chain Rule

Given that 𝑦=49𝑥5tan, determine 𝑦.

Answer

To find 𝑦, we need to differentiate 𝑦 with respect to 𝑥 twice. We will start by finding the first derivative by recalling the following trigonometric derivative result. For any constant 𝑘, ddtansec𝑥(𝑘𝑥)=𝑘𝑘𝑥.

Setting 𝑘=95 and taking the constant factor of 4 outside of the derivative gives us 𝑦=4𝑥9𝑥5=4959𝑥5=3659𝑥5.ddtansecsec

We need to differentiate this once more to find the second derivative. Since this is a composition of differentiable functions, we will differentiate this function by using the chain rule:

If 𝑢 is differentiable at 𝑥 and 𝑣 is differentiable at 𝑢(𝑥), then dddddd𝑥(𝑢(𝑣(𝑥)))=𝑢𝑣𝑣𝑥.

Setting 𝑢(𝑣)=𝑣 and 𝑣(𝑥)=9𝑥5sec gives us 𝑦=365𝑢(𝑣(𝑥)).

To use the chain rule, we need to differentiate 𝑢 and 𝑣 using the power rule for differentiation, and for any constant 𝑘, ddsecsectanddddsectan𝑥(𝑘𝑥)=𝑘𝑘𝑥𝑘𝑥,𝑢𝑣=2𝑣,𝑣𝑥=959𝑥59𝑥5.

Substituting these into the chain rule formula, we get 𝑦=365𝑢𝑣𝑣𝑥=3652𝑣959𝑥59𝑥5=64825𝑣9𝑥59𝑥5.ddddsectansectan

Finally, we substitute 𝑣=9𝑥5sec to get 𝑦=648259𝑥59𝑥5.sectan

Let’s finish by recapping some basic points.

Key Points

  • We can differentiate functions multiple times to find higher-order derivatives.
  • The 𝑛th derivative of a function 𝑦=𝑓(𝑥) can be written as
    • dd𝑦𝑥,
    • 𝑓(𝑥)(),
    • 𝑓(𝑥), where the prime symbol appears 𝑛 times.
  • We can represent evaluating the derivative of a function 𝑦=𝑓(𝑥) using two different notations. For example, for the second derivative at 𝑥=𝑎, we can write
    • 𝑦𝑥dd,
    • 𝑓(𝑎).

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.