115 LearnersLast updated on August 5th, 2025
Derivative of Sinh
We use the derivative of sinh(x), which is cosh(x), to understand how the hyperbolic sine function changes in response to a slight change in x. Derivatives help us calculate various rates of change in real-life situations. We will now discuss the derivative of sinh(x) in detail.
What is the Derivative of Sinh?
We now understand the derivative of sinh(x). It is commonly represented as d/dx (sinh x) or (sinh x)', and its value is cosh(x).
The function sinh(x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below: Hyperbolic Sine Function: sinh(x) = (e^x - e^-x)/2.
Exponential Function: The derivative of exponential functions is crucial for differentiating sinh(x).
Hyperbolic Cosine Function: cosh(x) = (e^x + e^-x)/2.
Derivative of Sinh Formula
The derivative of sinh(x) can be denoted as d/dx (sinh x) or (sinh x)'.
The formula we use to differentiate sinh(x) is: d/dx (sinh x) = cosh x The formula applies to all x.
Proofs of the Derivative of Sinh
We can derive the derivative of sinh(x) using proofs. To show this, we will use the properties of exponential functions along with the rules of differentiation.
There are several methods we use to prove this, such as: Using Exponential Definitions Using Limits
We will now demonstrate that the differentiation of sinh(x) results in cosh(x) using the above-mentioned methods: Using Exponential Definitions The derivative of sinh(x) can be proved using the definition of sinh(x) in terms of exponential functions: sinh(x) = (e^x - e^-x)/2
By differentiating, we apply the derivative of exponentials: d/dx [sinh(x)] = d/dx [(e^x - e^-x)/2] = (1/2) [d/dx (e^x) - d/dx (e^-x)] = (1/2) [e^x + e^-x] = cosh(x) Hence, proved.
Using Limits The derivative of sinh(x) can also be derived using limits. Consider: f(x) = sinh(x) f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [(sinh(x + h) - sinh(x)) / h] = limₕ→₀ [((e^(x+h) - e^-(x+h)) - (e^x - e^-x)) / (2h)] = limₕ→₀ [(e^x(e^h - 1) - e^-x(e^h - 1)) / (2h)] = (1/2) [e^x - e^-x] = cosh(x) Hence, proved.
Higher-Order Derivatives of Sinh
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 sinh(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 sinh(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).
Special Cases:
When x is 0, the derivative of sinh(x) = cosh(0), which is 1.
Common Mistakes and How to Avoid Them in Derivatives of Sinh
Students frequently make mistakes when differentiating sinh(x).
These mistakes can be resolved by understanding the proper solutions.
Here are a few common mistakes and ways to solve them:
Mistake 1
Not simplifying the equation
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 using exponential identities.
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
Forgetting the Basic Definitions
They might not remember the basic definitions of hyperbolic functions, such as sinh(x) = (e^x - e^-x)/2.
Keep in mind that these definitions are crucial for understanding and differentiating hyperbolic functions properly.
Mistake 3
Incorrect use of Exponential Rules
While differentiating functions such as sinh(x)/x, students misapply the rules of exponentials.
For example, incorrect differentiation: d/dx (sinh(x) / x) = cosh(x)/x². d/dx (u/v) = (v . u’ - u . v’)/ v² (where u = sinh(x) and v = x) Applying the quotient rule, d/dx (sinh(x)/x) = (x . cosh(x) - sinh(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
Not writing Constants and Coefficients
There is a common mistake that students sometimes forget to multiply the constants placed before sinh(x).
For example, they incorrectly write d/dx (5 sinh(x)) = cosh(x).
Students should check the constants in the terms and ensure they are multiplied properly.
For example, the correct equation is d/dx (5 sinh(x)) = 5 cosh(x).
Mistake 5
Not Applying the Chain Rule
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 (sinh(2x)) = cosh(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 (sinh(2x)) = 2 cosh(2x).
Examples Using the Derivative of Sinh
Problem 1
Calculate the derivative of (sinh(x)·cosh(x))
Here, we have f(x) = sinh(x)·cosh(x). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sinh(x) and v = cosh(x).
Let’s differentiate each term, u′ = d/dx (sinh(x)) = cosh(x) v′ = d/dx (cosh(x)) = sinh(x) substituting into the given equation, f'(x) = (cosh(x)).(cosh(x)) + (sinh(x)).(sinh(x))
Let’s simplify terms to get the final answer, f'(x) = cosh²(x) + sinh²(x) Thus, the derivative of the specified function is cosh²(x) + sinh²(x).
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.
Problem 2
A construction company is building a bridge, and the elevation is represented by the function y = sinh(x) where y represents the height of the bridge at a distance x. If x = 1 meter, measure the slope of the bridge.
We have y = sinh(x) (slope of the bridge)...(1) Now, we will differentiate the equation (1) Take the derivative sinh(x): dy/dx = cosh(x) Given x = 1 (substitute this into the derivative) dy/dx = cosh(1)
Hence, we get the slope of the bridge at a distance x = 1 as cosh(1).
Explanation
We find the slope of the bridge at x = 1 as cosh(1), which provides the rate of change of the elevation at that point.
Problem 3
Derive the second derivative of the function y = sinh(x).
The first step is to find the first derivative, dy/dx = cosh(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cosh(x)] = sinh(x)
Therefore, the second derivative of the function y = sinh(x) is sinh(x).
Explanation
We use the step-by-step process, where we start with the first derivative.
Using the derivative of cosh(x), we find the second derivative, which simplifies to sinh(x).
Problem 4
Prove: d/dx (sinh²(x)) = 2 sinh(x) cosh(x).
Let’s start using the chain rule: Consider y = sinh²(x) = [sinh(x)]² To differentiate, we use the chain rule: dy/dx = 2 sinh(x). d/dx [sinh(x)]
Since the derivative of sinh(x) is cosh(x), dy/dx = 2 sinh(x)·cosh(x)
Substituting y = sinh²(x), d/dx (sinh²(x)) = 2 sinh(x)·cosh(x) Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sinh(x) with its derivative. As a final step, we substitute y = sinh²(x) to derive the equation.
Problem 5
Solve: d/dx (sinh(x)/x)
To differentiate the function, we use the quotient rule: d/dx (sinh(x)/x) = (d/dx (sinh(x)).x - sinh(x).d/dx(x))/x²
We will substitute d/dx (sinh(x)) = cosh(x) and d/dx(x) = 1 = (x·cosh(x) - sinh(x))/x²
Therefore, d/dx (sinh(x)/x) = (x·cosh(x) - sinh(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.
FAQs on the Derivative of Sinh
1.Find the derivative of sinh(x).
2.Can we use the derivative of sinh(x) in real life?
3.Is the derivative of sinh(x) always defined?
4.What rule is used to differentiate sinh(x)/x?
5.Are the derivatives of sinh(x) and sinh⁻¹(x) the same?
Important Glossaries for the Derivative of Sinh
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Hyperbolic Sine Function: A hyperbolic function represented as sinh(x) = (e^x - e^-x)/2.
- Hyperbolic Cosine Function: A hyperbolic function represented as cosh(x) = (e^x + e^-x)/2.
- Exponential Function: A function in which an independent variable is in the exponent; crucial for differentiating hyperbolic functions.
- Chain Rule: A rule for finding the derivative of a composition of functions.
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Jaskaran Singh Saluja
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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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