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Lesson Explainer: Differentiation of Exponential Functions Mathematics • Higher Education

In this explainer, we will learn how to find the derivatives of exponential functions.

Exponential functions play an important role in mathematics and have many physical applications. For example, we use exponential functions to model the growth of bacteria, continuously compounded interest, and radioactive decay, to mention just a few of its applications. In fact, its ubiquitous use in mathematics led an Austrian American mathematician, Walter Rudin, to call it the most important function in mathematics. One of the reasons why exponential functions are so important in mathematics is their properties in relation to derivatives; in this explainer, we will explore these properties. We begin with a general definition of an exponential function.

Definition: Exponential Function

An exponential function is a function of the form 𝑓(𝑥)=𝑏, where 𝑏 is a constant satisfying 𝑏>0 and 𝑏1. 𝑏 is called the base of the exponential function.

A common misconception when it comes to derivatives of exponential functions is to assume that we can apply the power rule ddforanyconstant𝑥𝑥=𝑝𝑥,𝑝.

This is not true as we shall see. It is important to be clear on the distinction: the power rule applies when the base of the exponent is the variable and the exponent is fixed, whereas for exponential functions, the exponent is the variable and the base is fixed.

Power FunctionsExponential Functions
𝑥𝑏
The variable is the base.The base is constant.
The exponent is constant.The variable is the exponent.

We can differentiate power functions by using the power rule. For any 𝑝, dd𝑥(𝑥)=𝑝𝑥.

However, the power rule can only apply to power functions and not to exponential functions. Application of the power rule to differentiate an exponential function is a common mistake in differentiation. For instance, applying the power rule to find the derivative of 2 would result in 𝑥2, which is incorrect. We need to be careful to recognize that the variable in 2 is in the exponent, which makes it an exponential function. Hence, we cannot apply the power rule to differentiate this exponential function.

Let us consider how to differentiate an exponential function. We will begin by finding the derivative of the natural exponential function, 𝑒, whose base 𝑒 is known as Euler’s constant. Euler’s constant is defined through the following limit:

𝑒=1+1𝑛.lim(1)

After finding the derivative of the natural exponential function, we will learn how to differentiate general exponential functions, which are in the form 𝑏 for some 𝑏>0 and 𝑏1.

Recall the derivative of a function 𝑓(𝑥) is given by ddlim𝑥𝑓(𝑥)=𝑓(𝑥+)𝑓(𝑥).

Applying this definition to the natural exponential function 𝑓(𝑥)=𝑒 gives us ddlim𝑥(𝑒)=𝑒𝑒.

Using the rule of exponents, we can write 𝑒 as 𝑒×𝑒. Then, this limit is written as lim𝑒×𝑒𝑒.

In the numerator of the quotient, we note that 𝑒 is a common factor. Hence, the factorized form is given by lim𝑒×𝑒1.

Since 𝑒 does not depend on , we can treat it as a constant in regard to the limit as 0. Hence, we can factor out this expression from the limit to write

ddlim𝑥(𝑒)=𝑒𝑒1.(2)

Hence, evaluating the limit lim𝑒1 will lead to the formula for the derivative of 𝑒. Let us find this limit.

Computing this limit involves a few tricks, which are not apparent at first. But it will be helpful to see how exactly the definition of Euler’s constant given in (1) impacts the computation of this limit.

We begin by defining variable 𝑦 to equal the numerator of the quotient within the limit: 𝑦=𝑒1.

We can rearrange this equation so that is the subject: 𝑦+1=𝑒=(𝑦+1).ln

We can substitute the numerator of the quotient by 𝑦 and the denominator of the quotient by ln(𝑦+1). We also note that when approaches 0, the variable 𝑦 also approaches 0 since lim𝑒1=𝑒1=0. Changing the variable of the limit from to 𝑦, we can write limlimln𝑒1=𝑦(𝑦+1).

Multiplying the numerator and denominator by 1𝑦, we can write limlnlimln𝑦×(𝑦+1)=1(𝑦+1).

Recall the law of logarithms that tells us that the coefficient of a log becomes the exponent of its argument. Hence, applying the concepts of continuous functions, we can write the limit above as

limlnlnlim1(𝑦+1)=1(𝑦+1).(3)

Now, consider the limit within the parenthesis above. This resembles somewhat the limit in equation (1), which is the definition of Euler’s constant. Let us define another variable 𝑛 by 𝑛=1𝑦.

This means also that 𝑦=1𝑛. We note that as 𝑦 approaches 0, 𝑛 approaches infinity. This leads to limlim(𝑦+1)=1+1𝑛.

By equation (1), the limit on the right-hand side is equal to 𝑒. Substituting the value of this limit in (3), we can find the value of the desired limit: limln𝑒1=1𝑒=1.

Now that we have found the value of this limit, we can return to differentiating 𝑒. Substituting this limit value in equation (2), we have ddlim𝑥(𝑒)=𝑒×𝑒1=𝑒.

This gives us the derivative of the natural exponential function.

Rule: Derivative of the Natural Exponential Function

The derivative of the natural exponential function is dd𝑥(𝑒)=𝑒.

The significance of the derivative rule for the natural exponential function cannot be overstated. The natural exponential function is the only nonzero function whose derivative is equal to itself. This fact makes the natural exponential function appear as a solution for many different mathematical models of real-world problems. Beyond any doubt, the natural exponential function is the most prominent function in mathematical models of the real world.

Let us begin with an example where we apply this rule to differentiate a given function.

Example 1: Differentiating Functions Involving Exponentials

Differentiate the function 𝑦=3𝑒5𝑥.

Answer

In this example, we need to differentiate a function that involves the exponential function 𝑒 and a root function 5𝑥. We note that the given root function can be expressed as a power function by 5𝑥=5𝑥=5𝑥.

Using this form, we can rewrite the given function as 𝑦=3𝑒5𝑥.

Hence, the derivative is written as 𝑦=3𝑒5𝑥.

We recall the sum/difference rule that allows us to split up the sum inside the derivative and also the constant multiple rule that allows us to factor the constants 3 and 5 outside the derivatives. Hence, 𝑦=(3𝑒)5𝑥=3(𝑒)5𝑥.

Now, we need to differentiate the exponential and power functions. We recall the rules of differentiation for these functions: (𝑥)=𝑝𝑥,𝑝,(𝑒)=𝑒.foranyconstant

Applying these rules to 𝑦, we obtain 𝑦=3𝑒513𝑥=3𝑒+53𝑥.

Hence, 𝑦=3𝑒+53𝑥.

In the next example, we will use the derivative of the natural exponential function multiplied to a reciprocal trigonometric function.

Example 2: Finding the Derivative of a Function Involving Trigonometric and Exponential Functions Using the Product Rule

Differentiate 𝑓(𝑥)=𝑒𝑥sec.

Answer

We can begin by noticing that the given function is a product of two functions, 𝑒 and sec𝑥. We recall the product rule: given differentiable functions 𝑔(𝑥) and (𝑥), (𝑔(𝑥)(𝑥))=𝑔(𝑥)(𝑥)+𝑔(𝑥)(𝑥).

Applying the product rule to our function, 𝑓(𝑥)=(𝑒)𝑥+𝑒(𝑥).secsec

Now, we need to compute the derivatives (𝑒) and (𝑥)sec. We recall the derivatives of the exponential and reciprocal trigonometric functions: (𝑒)=𝑒,(𝑥)=𝑥𝑥.secsectan

Substituting these expressions into 𝑓(𝑥), we have 𝑓(𝑥)=𝑒𝑥+𝑒𝑥𝑥.secsectan

We can see that 𝑒𝑥sec is a common factor of both terms on the right-hand side of the equation. Factoring out this term, we have 𝑓(𝑥)=𝑒𝑥(1+𝑥).sectan

In our next example, we will apply the derivative rule for the natural exponential function combined with the chain rule to differentiate a given function.

Example 3: Differentiating Exponential Functions Using the Chain Rule

If 𝑓(𝑥)=5𝑒, find 𝑓(𝑥).

Answer

In this example, we need to find 𝑓(𝑥)=5𝑒.

Before we differentiate the exponential function, we can factor out the constant 5 outside the derivative using the constant multiple rule: 𝑓(𝑥)=5𝑒=5𝑒.

Now, we consider the derivative 𝑒. We can note that 𝑒 is a composition of two functions; hence, we recall the chain rule for differentiating a composition of functions: (𝑔((𝑥)))=𝑔((𝑥))(𝑥).

For the function 𝑒, the outside function is 𝑔(𝑢)=𝑒 and the inside function is (𝑥)=9𝑥. We need to compute the derivatives of each of these functions. To compute the derivative of 𝑔(𝑢)=𝑒, we recall the rule for differentiating the exponential function, (𝑒)=𝑒.

This leads to 𝑔(𝑢)=(𝑒)=𝑒.

On the other hand, we can apply the power rule to differentiate (𝑥)=(9𝑥)=9.

Substituting these expressions into the chain rule, we have 𝑒=𝑒×(9)=9𝑒.

Finally, remembering that we have factored out 5 in the beginning, we can write 𝑓(𝑥)=59𝑒=45𝑒.

In the previous example, we applied the chain rule to differentiate an exponential function. Using the same approach, we can write a more general formula for chain rules that involve the exponential function.

Rule: Chain Rule for the Natural Exponential Function

Given a differentiable function 𝑢(𝑥), we have 𝑒=𝑢(𝑥)𝑒.()()

When we differentiated 𝑒 in the previous example, we could have noted that this is the chain rule with the exponential function where 𝑢(𝑥)=9𝑥. Since 𝑢(𝑥)=9, this leads to the derivative 𝑒=9𝑒.

This is not a formula that must be memorized since we can always find this derivative by writing out the chain rule as we did in the first example. However, knowing this extra rule will often save our time when we differentiate these functions.

In the next example, we will apply this rule to differentiate an exponential function with a quadratic exponent.

Example 4: Differentiating Natural Exponential Functions Using the Chain Rule

Determine the derivative of 𝑔(𝑥)=5𝑒.

Answer

In this example, we need to find the derivative 𝑔(𝑥)=5𝑒.

We can factor out the constant 5 outside the derivative using the constant multiple rule: 𝑔(𝑥)=5𝑒.

To compute the derivative 𝑒, we note that this is a composition of functions in the form 𝑒(), where 𝑢(𝑥)=3𝑥+3𝑥. Recall the chain rule for the natural exponential function: 𝑒=𝑒𝑢(𝑥).()()

Using the power rule, we can compute 𝑢(𝑥)=6𝑥+3.

Substituting this derivative into the chain rule, we have 𝑒=𝑒(6𝑥+3).

Finally, remembering that we have factored out 5 in the beginning, we can write 𝑔(𝑥)=5𝑒(6𝑥+3)=𝑒(30𝑥15).

Hence, 𝑔(𝑥)=𝑒(30𝑥15).

In the next example, we will apply the chain rule for the natural exponential function with the quotient rule to differentiate a given function.

Example 5: Differentiating Combinations of Exponential and Polynomial Functions Using the Quotient Rule

Find the first derivative of the function 𝑦=4𝑒7𝑥+4.

Answer

We can begin by noticing that the given function is a quotient of two functions, 4𝑒 and 7𝑥+4. Hence, we recall the quotient rule: given two differentiable functions 𝑔(𝑥) and (𝑥), 𝑔(𝑥)(𝑥)=𝑔(𝑥)(𝑥)𝑔(𝑥)(𝑥)((𝑥)).

Applying the quotient rule to our function, we have 𝑦=4𝑒(7𝑥+4)4𝑒(7𝑥+4)(7𝑥+4).

We now need to compute the derivatives 4𝑒 and (7𝑥+4). The derivative (7𝑥+4)=7 by applying the power rule. For the other derivative, we can first apply the constant multiple rule to factor out the constant 4 from the derivative: 4𝑒=4𝑒.

To compute 𝑒, we recall the chain rule for the natural exponential function: 𝑒=𝑒𝑢(𝑥).()()

In this case, 𝑢(𝑥)=7𝑥; hence, 𝑢(𝑥)=7 by the power rule. Substituting these expressions into the chain rule, 𝑒=𝑒×7, which means 4𝑒=4𝑒×7=28𝑒.

We can now substitute these derivatives into 𝑦 to obtain 𝑦=28𝑒(7𝑥+4)4𝑒(7)(7𝑥+4)=196𝑥𝑒+112𝑒28𝑒(7𝑥+4)=196𝑥𝑒+84𝑒(7𝑥+4).

So far, we have considered the rule for differentiating the natural exponential function. We now turn our attention to the general exponential function 𝑏, where the base 𝑏 satisfies 𝑏>0 and 𝑏1. We can relate a general exponential function to the natural exponential function when we use the identity 𝐵=𝐴,𝐴>0,𝐵>0.logforanyconstants

If we apply this identity with 𝐵=𝑏 and 𝐴=𝑒, then writing the log-based 𝑒 as ln (natural log), 𝑏=𝑒.ln

Recall the property of logarithm that states lnln𝑏=𝑥𝑏. This leads to 𝑏=𝑒.ln

Hence, a general exponential function can be written in form of a natural exponential function. To find the derivative of a general exponential function, we can apply the chain rule for the natural exponential function with 𝑢(𝑥)=𝑥𝑏ln. We can apply the power rule to obtain 𝑢(𝑥)=𝑏ln, so we can write (𝑏)=𝑒𝑏.lnln

Since 𝑒=𝑏ln, we can substitute this expression to obtain the derivative of a general exponential function.

Rule: Derivative of General Exponential Functions

The derivative of an exponential function 𝑏, if 𝑏>0 and 𝑏1, is ddln𝑥(𝑏)=𝑏𝑏.

Let us consider an example where we use the derivative of a general exponential function to find the derivative of a given function.

Example 6: Finding the First Derivative of an Exponential Function with an Integer Base

If 𝑦=3×2, determine dd𝑦𝑥.

Answer

In this example, we need to determine dddd𝑦𝑥=𝑥(3×2).

We can factor out the constant 3 outside the derivative using the constant multiple rule: dddd𝑦𝑥=3𝑥(2).

To compute the derivative of the exponential function 2, we recall the derivative of an exponential function: ddln𝑥(𝑏)=𝑏𝑏.

Since we have the base 𝑏=2 for our exponential function, we can write ddln𝑥(2)=22.

Substituting this derivative into dd𝑦𝑥, we can obtain ddlnln𝑦𝑥=3(22)=3×22.

In our next example, we will apply the chain rule with the derivative of a general exponential function. While we have established a formula for the chain rule for the natural exponential function, this formula does not directly lead to the chain rule for general exponential functions. In an effort to avoid over-producing formulae, we will compute these derivatives by applying the chain rule directly.

Example 7: Differentiating Exponential Functions Using the Chain Rule

Find the first derivative of the function 𝑦=7.

Answer

In this example, we need to find the first derivative of an exponential function that is raised to another exponent. Although this expression looks complex, we can use the rules of exponents to significantly simplify the expression. We recall that, for any base 𝑏 and exponents 𝑐 and 𝑑, (𝑏)=𝑏.

In other words, we can multiply these exponents together. Using this rule, we can simplify the given function as 𝑦=7=7.()

Now, we compute the first derivative: 𝑦=7.

We can note that 7 is a composition of two functions; hence, we recall the chain rule for differentiating a composition of functions: (𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

For the function 7, the outside function is 𝑓(𝑢)=7 and the inside function is 𝑔(𝑥)=18𝑥+16. We need to compute the derivatives of each of these functions. To compute the derivative of 𝑓(𝑢)=7, we recall the rule for differentiating general exponential functions, (𝑏)=𝑏𝑏,𝑏>0,𝑏1.lnforanyconstant

This leads to 𝑓(𝑢)=(7)=77.ln

On the other hand, we can apply the power rule to differentiate 𝑔(𝑥)=(18𝑥+16)=18.

Substituting these expressions into the chain rule, we have 𝑦=77×18=1877.lnln

In the previous example, we applied the chain rule for a general exponential function to differentiate a given function. We can follow the same process to find a general formula for the chain rule for general exponential functions.

Rule: Chain Rule for General Exponential Functions

Given a differentiable function 𝑢(𝑥) and a constant 𝑏 satisfying 𝑏>0 and 𝑏1, we have 𝑏=𝑢(𝑥)𝑏𝑏.()()ln

In our final example, we will consider a real-world problem involving the derivative of the natural exponential function.

Example 8: Finding the Rate of Change of Exponential Functions in a Real-World Context

A factory’s production of 𝑦 units in a day 𝑡 is governed by the relation 𝑦=40010𝑒. What is the rate of change of production with respect to time on the fifth day?

Answer

We recall that the rate of change of a function is given by the derivative of the function evaluated at a point. In this example, our function represents a factory’s production. To find the rate of change of production with respect to time on the fifth day, we need to find the derivative of this function and evaluate it at 𝑡=5.

Let us find the derivative dd𝑦𝑡. We can apply the constant multiple rule and sum/difference rule to write dddddddddd𝑦𝑡=𝑡40010𝑒=400𝑡10𝑒=400𝑡(10)𝑡𝑒.

We now need to compute the derivatives dd𝑡(10) and dd𝑡𝑒. By the constant rule, we know that dd𝑡(10)=0.

To compute the derivative dd𝑡𝑒, we recall the chain rule for the natural exponential function: given a differentiable function 𝑢(𝑡), dd𝑡𝑒=𝑢(𝑡)𝑒.()()

Here, 𝑢(𝑡)=0.8𝑡; hence, we have 𝑢(𝑡)=0.8 by the power rule. Substituting these derivatives in the chain rule, dd𝑡𝑒=0.8𝑒.

We can now substitute these derivatives into dd𝑦𝑡 to write dd𝑦𝑡=40000.8𝑒=320𝑒.

Evaluating the derivative at 𝑡=5, dd𝑦𝑡|||=320𝑒=320𝑒.×

Hence, the rate of change of production with respect to time on the fifth day is 320𝑒.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • The derivative of the natural exponential function is dd𝑥(𝑒)=𝑒.
  • Given a differentiable function 𝑢(𝑥), we have 𝑒=𝑢(𝑥)𝑒.()()
  • The derivative of an exponential function 𝑏, if 𝑏>0 and 𝑏1, is ddln𝑥(𝑏)=𝑏𝑏.
  • Given a differentiable function 𝑢(𝑥) and a constant 𝑏 satisfying 𝑏>0 and 𝑏1, we have 𝑏=𝑢(𝑥)𝑏𝑏.()()ln

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