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Last updated on October 17, 2025

Derivative of sech x

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We use the derivative of sech(x), which is -sech(x)tanh(x), as a measuring tool for how the hyperbolic secant function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of sech(x) in detail.

Derivative of sech x for US Students
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What is the Derivative of sech x?

We now understand the derivative of sech x. It is commonly represented as d/dx (sech x) or (sech x)', and its value is -sech(x)tanh(x). The function sech x has a clearly defined derivative, indicating it is differentiable within its domain.

 

The key concepts are mentioned below:

 

Hyperbolic Secant Function: sech(x) = 1/cosh(x).

 

Quotient Rule: Rule for differentiating sech(x) (since it consists of 1/cosh(x)).

 

Hyperbolic Tangent Function: tanh(x) = sinh(x)/cosh(x).

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Derivative of sech x Formula

The derivative of sech x can be denoted as d/dx (sech x) or (sech x)'. The formula we use to differentiate sech x is: d/dx (sech x) = -sech(x)tanh(x) (sech x)' = -sech(x)tanh(x) The formula applies to all x where cosh(x) ≠ 0.

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Proofs of the Derivative of sech x

We can derive the derivative of sech x using proofs. To show this, we will use the hyperbolic identities along with the rules of differentiation.

 

There are several methods we use to prove this, such as:

 

  • By First Principle
     
  • Using Chain Rule
     
  • Using Quotient Rule

 

We will now demonstrate that the differentiation of sech x results in -sech(x)tanh(x) using the above-mentioned methods:

 

By First Principle

 

The derivative of sech x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of sech x using the first principle, we will consider f(x) = sech x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = sech x, we write f(x + h) = sech (x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [sech(x + h) - sech x] / h Using the definition of sech and trigonometric identities, we can simplify it to show that: f'(x) = -sech(x)tanh(x). Hence, proved.

 

Using Chain Rule

 

To prove the differentiation of sech x using the chain rule, We use the formula: sech x = 1/ cosh x Consider f(x) = 1 and g (x)= cosh x So we get, sech x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]²… (1) Let’s substitute f(x) = 1 and g(x) = cosh x in equation (1), d/dx (sech x) = [0(cosh x) - 1(sinh x)] / (cosh x)² = -sinh x / (cosh x)² = -sech(x)tanh(x) Since tanh x = sinh x/cosh x, we write: d/dx(sech x) = -sech(x)tanh(x).

 

Using Quotient Rule

 

We will now prove the derivative of sech x using the quotient rule. The step-by-step process is demonstrated below: Here, we use the formula, sech x = 1/cosh x Given that, u = 1 and v = cosh x Using the quotient rule formula: d/dx [u/v] = [u'v - uv'] / v² u' = d/dx (1) = 0 (substitute u = 1) v' = d/dx (cosh x) = sinh x Using the quotient rule formula: d/dx (sech x) = (0 . v - 1 . sinh x) / (cosh x)² = -sinh x / (cosh x)² Since -sinh x / (cosh x)² = -sech(x)tanh(x), we have: d/dx (sech x) = -sech(x)tanh(x).

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Higher-Order Derivatives of sech x

When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like sech(x).

 

For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.

 

For the nth Derivative of sech(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).

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Special Cases:

When x approaches infinity, the derivative approaches zero because sech(x) tends to zero.

 

When x is zero, the derivative of sech x = -sech(x)tanh(x), which is 0 because tanh(0)=0.

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Common Mistakes and How to Avoid Them in Derivatives of sech x

Students frequently make mistakes when differentiating sech x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Mistake 1

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Not simplifying the equation

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Students may forget to simplify the equation, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result, especially when solving using the product or quotient rule.

 

Ensure that each step is written in order. Students might think it is awkward, but it is important to avoid errors in the process.

Mistake 2

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Forgetting the Undefined Points of sech x

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They might not remember that sech x is undefined for complex values where cosh x equals zero. Keep in mind that you should consider the domain of the function that you differentiate.

 

It will help you understand that the function is not continuous at such certain points.

Mistake 3

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Incorrect use of Quotient Rule

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While differentiating functions such as sech x/x, students misapply the quotient rule. For example: Incorrect differentiation: d/dx (sech x / x) = -sech(x)tanh(x)/x². d/dx (u/v) = (v . u’ - u . v’) / v² (where u = sech x and v = x)

 

Applying the quotient rule, d/dx (sech x/x) = (x(-sech(x)tanh(x)) - sech x)/x² To avoid this mistake, write the quotient rule without errors. Always check for errors in the calculation and ensure it is properly simplified.

Mistake 4

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Not writing Constants and Coefficients

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There is a common mistake that students at times forget to multiply the constants placed before sech x. For example, they incorrectly write d/dx (5 sech x) = -sech(x)tanh(x).

 

Students should check the constants in the terms and ensure they are multiplied properly. For e.g., the correct equation is d/dx (5 sech x) = 5(-sech(x)tanh(x)).

Mistake 5

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Not Applying the Chain Rule

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Students often forget to use the chain rule. This happens when the derivative of the inner function is not considered. For example: Incorrect: d/dx (sech(2x)) = -sech(2x)tanh(2x).

 

To fix this error, students should divide the functions into inner and outer parts. Then, make sure that each function is differentiated. For example, d/dx (sech(2x)) = -2 sech(2x)tanh(2x).

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Examples Using the Derivative of sech x

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Problem 1

Calculate the derivative of (sech x·tanh x)

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Here, we have f(x) = sech x·tanh x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sech x and v = tanh x. Let’s differentiate each term, u′ = d/dx (sech x) = -sech(x)tanh(x) v′ = d/dx (tanh x) = sech²x substituting into the given equation, f'(x) = (-sech(x)tanh(x)).(tanh x) + (sech x).(sech²x) Let’s simplify terms to get the final answer, f'(x) = -sech(x)tanh²x + sech³x Thus, the derivative of the specified function is -sech(x)tanh²x + sech³x.

Explanation

We find the derivative of the given function by dividing the function into two parts.

The first step is finding its derivative and then combining them using the product rule to get the final result.

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Problem 2

A scientist is modeling the decay of a chemical with concentration represented by C(x) = sech(x), where x is time in seconds. After 3 seconds, find the rate of change of concentration.

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We have C(x) = sech(x) (concentration of the chemical)...(1) Now, we will differentiate the equation (1) Take the derivative sech(x): dC/dx = -sech(x)tanh(x) Given x = 3 (substitute this into the derivative) dC/dx at x = 3 = -sech(3)tanh(3) We need to calculate this numerically to get the specific value. Hence, we get the rate of change of concentration at x = 3.

Explanation

We find the rate of change of concentration at x = 3, which means how rapidly the concentration of the chemical changes after 3 seconds.

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Problem 3

Derive the second derivative of the function y = sech(x).

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The first step is to find the first derivative, dy/dx = -sech(x)tanh(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-sech(x)tanh(x)] Here we use the product rule, d²y/dx² = -[sech'(x)tanh(x) + sech(x)tanh'(x)] = -[(-sech(x)tanh(x))tanh(x) + sech(x)(sech²x)] = sech(x)tanh²(x) - sech³(x) Therefore, the second derivative of the function y = sech(x) is sech(x)tanh²(x) - sech³(x).

Explanation

We use the step-by-step process, where we start with the first derivative.

Using the product rule, we differentiate -sech(x)tanh(x).

We then simplify the terms to find the final answer.

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Problem 4

Prove: d/dx (sech²(x)) = -2 sech²(x)tanh(x).

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Let’s start using the chain rule: Consider y = sech²(x) [sech(x)]² To differentiate, we use the chain rule: dy/dx = 2 sech(x) d/dx [sech(x)] Since the derivative of sech(x) is -sech(x)tanh(x), dy/dx = 2 sech(x)(-sech(x)tanh(x)) Substituting y = sech²(x), d/dx (sech²(x)) = -2 sech²(x)tanh(x) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation.

Then, we replace sech(x) with its derivative.

As a final step, we substitute y = sech²(x) to derive the equation.

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Problem 5

Solve: d/dx (sech x/x)

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To differentiate the function, we use the quotient rule: d/dx (sech x/x) = (d/dx (sech x)·x - sech x·d/dx(x)) / x² We will substitute d/dx (sech x) = -sech(x)tanh(x) and d/dx(x) = 1 = [x(-sech(x)tanh(x)) - sech x] / x² = [-x sech(x)tanh(x) - sech x] / x² Therefore, d/dx (sech x/x) = [-x sech(x)tanh(x) - sech x] / x²

Explanation

In this process, we differentiate the given function using the quotient rule.

As a final step, we simplify the equation to obtain the final result.

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FAQs on the Derivative of sech x

1.Find the derivative of sech x.

Using the quotient rule on sech x gives 1/cosh x, d/dx (sech x) = -sech(x)tanh(x) (simplified)

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2.Can we use the derivative of sech x in real life?

Yes, we can use the derivative of sech x in real life in modeling decay processes or wave functions in physics and engineering.

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3.Is it possible to take the derivative of sech x at the point where x = π/2?

Yes, sech x is defined at x = π/2. The derivative can be calculated without issue at this point.

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4.What rule is used to differentiate sech x/x?

We use the quotient rule to differentiate sech x/x, d/dx (sech x/x) = [-x sech(x)tanh(x) - sech x] / x².

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5.Are the derivatives of sech x and sech⁻¹x the same?

No, they are different. The derivative of sech x is -sech(x)tanh(x), while the derivative of sech⁻¹x involves a more complex expression: -1/(x√(1-x²)) for |x|<1.

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Important Glossaries for the Derivative of sech x

  • Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.

 

  • Hyperbolic Secant Function: A hyperbolic function represented as sech x, the reciprocal of the hyperbolic cosine.

 

  • Hyperbolic Tangent Function: A hyperbolic function represented as tanh x, the ratio of sinh x to cosh x.

 

  • Quotient Rule: A method for finding the derivative of a quotient of two functions.

 

  • Chain Rule: A formula for computing the derivative of the composition of two or more functions.
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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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