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104 LearnersLast updated on September 15, 2025
Derivative of 5secx
We use the derivative of 5sec(x), which is 5sec(x)tan(x), as a measuring tool for how the 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 5sec(x) in detail.
What is the Derivative of 5secx?
We now understand the derivative of 5secx. It is commonly represented as d/dx (5secx) or (5secx)', and its value is 5sec(x)tan(x). The function 5secx has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Secant Function: sec(x) = 1/cos(x).
Product Rule: Rule for differentiating products of functions.
Derivative of Secant: The derivative of sec(x) is sec(x)tan(x).
Derivative of 5secx Formula
The derivative of 5secx can be denoted as d/dx (5secx) or (5secx)'.
The formula we use to differentiate 5secx is: d/dx (5secx) = 5sec(x)tan(x)
The formula applies to all x where cos(x) ≠ 0.
Proofs of the Derivative of 5secx
We can derive the derivative of 5secx using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:
- Using Chain Rule
- Using Product Rule
We will now demonstrate that the differentiation of 5secx results in 5sec(x)tan(x) using the above-mentioned methods:
Using Chain Rule
To prove the differentiation of 5secx using the chain rule, We use the formula: 5secx = 5 * sec(x) Let f(x) = sec(x)
By chain rule: d/dx [5 * f(x)] = 5 * f'(x)
The derivative of sec(x) is sec(x)tan(x), so f'(x) = sec(x)tan(x)
Thus, d/dx (5secx) = 5 * sec(x)tan(x)
Using Product Rule
We will now prove the derivative of 5secx using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, 5secx = 5 * sec(x)
Given that u = 5 and v = sec(x)
Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (5) = 0 v' = d/dx (sec(x)) = sec(x)tan(x)
So, d/dx (5secx) = 0 * sec(x) + 5 * sec(x)tan(x) = 5sec(x)tan(x)
Hence, proved.
Higher-Order Derivatives of 5secx
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 5sec(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 5sec(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 π/2, the derivative is undefined because sec(x) has a vertical asymptote there. When x is 0, the derivative of 5secx = 5sec(0)tan(0), which is 0.
Common Mistakes and How to Avoid Them in Derivatives of 5secx
Students frequently make mistakes when differentiating 5secx. 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 solving using the product or chain 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
Forgetting the Undefined Points of Sec x
They might not remember that sec x is undefined at the points such as (x = π/2, 3π/2,...). 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
Incorrect use of Product Rule
While differentiating functions such as 5sec(x), students misapply the product rule. For example: Incorrect differentiation: d/dx (5sec(x)) = 5sec(x)tan(x) without considering the product rule. Applying the product rule, d/dx (5sec(x)) = 5 * d/dx (sec(x)) = 5sec(x)tan(x) To avoid this mistake, write the product 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 at times forget to multiply the constants placed before sec x. For example, they incorrectly write d/dx (5sec x) = sec(x)tan(x). Students should check the constants in the terms and ensure they are multiplied properly. For e.g., the correct equation is d/dx (5sec x) = 5sec(x)tan(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 (5sec(2x)) = 5sec(2x)tan(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 (5sec(2x)) = 10sec(2x)tan(2x).
Examples Using the Derivative of 5secx
Problem 1
Calculate the derivative of (5sec(x)tan(x))
Here, we have f(x) = 5sec(x)tan(x).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 5sec(x) and v = tan(x).
Let’s differentiate each term, u′= d/dx (5sec(x)) = 5sec(x)tan(x) v′= d/dx (tan(x)) = sec²(x)
Substituting into the given equation, f'(x) = (5sec(x)tan(x))(tan(x)) + (5sec(x))(sec²(x))
Let’s simplify terms to get the final answer, f'(x) = 5sec(x)tan²(x) + 5sec³(x)
Thus, the derivative of the specified function is 5sec(x)tan²(x) + 5sec³(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 company's stock price is modeled by the function y = 5sec(x), where y represents the price at time x. If x = π/6 hours, calculate the rate of change of the stock price.
We have y = 5sec(x) (stock price model)...(1)
Now, we will differentiate the equation (1) Take the derivative of 5sec(x): dy/dx = 5sec(x)tan(x)
Given x = π/6 (substitute this into the derivative)
dy/dx = 5sec(π/6)tan(π/6) Since sec(π/6) = 2/√3 and tan(π/6) = 1/√3, dy/dx = 5 * (2/√3) * (1/√3) = 10/3
Hence, the rate of change of the stock price at x = π/6 is 10/3.
Explanation
We find the rate of change of the stock price at x = π/6 as 10/3, which indicates the stock price is increasing at this rate per unit time.
Problem 3
Derive the second derivative of the function y = 5sec(x).
The first step is to find the first derivative, dy/dx = 5sec(x)tan(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [5sec(x)tan(x)]
Here we use the product rule, d²y/dx² = 5[d/dx (sec(x)tan(x))] = 5[sec(x)(sec²(x)) + tan(x)sec(x)tan(x)] = 5[sec³(x) + sec(x)tan²(x)]
Therefore, the second derivative of the function y = 5sec(x) is 5[sec³(x) + sec(x)tan²(x)].
Explanation
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate sec(x)tan(x). We then simplify the terms to find the final answer.
Problem 4
Prove: d/dx (5sec²(x)) = 10sec(x)sec(x)tan(x).
Let’s start using the chain rule: Consider y = 5sec²(x) = 5[sec(x)]² T
o differentiate, we use the chain rule: dy/dx = 10sec(x) * d/dx [sec(x)]
Since the derivative of sec(x) is sec(x)tan(x), dy/dx = 10sec(x) * sec(x)tan(x) = 10sec²(x)tan(x) Substituting y = 5sec²(x), d/dx (5sec²(x)) = 10sec²(x)tan(x)
Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sec(x) with its derivative. As a final step, we substitute y = 5sec²(x) to derive the equation.
Problem 5
Solve: d/dx (5sec(x)/x)
To differentiate the function, we use the quotient rule: d/dx (5sec(x)/x) = (d/dx (5sec(x)) * x - 5sec(x) * d/dx(x))/x²
We will substitute d/dx (5sec(x)) = 5sec(x)tan(x) and d/dx (x) = 1 = (5sec(x)tan(x) * x - 5sec(x) * 1) / x² = (5xsec(x)tan(x) - 5sec(x)) / x²
Therefore, d/dx (5sec(x)/x) = (5xsec(x)tan(x) - 5sec(x)) / x²
Explanation
In this process, we differentiate the given function using the product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.
FAQs on the Derivative of 5secx
1.Find the derivative of 5secx.
2.Can we use the derivative of 5secx in real life?
3.Is it possible to take the derivative of 5secx at the point where x = π/2?
4.What rule is used to differentiate 5sec(x)/x?
5.Are the derivatives of 5secx and secx the same?
Important Glossaries for the Derivative of 5secx
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Secant Function: A trigonometric function that is the reciprocal of the cosine function. It is typically represented as sec(x).
- Product Rule: A rule used to differentiate the product of two functions.
- Chain Rule: A rule used for differentiating compositions of functions.
- Undefined Points: Points where a function does not exist, often due to division by zero or other discontinuities.
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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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: He loves to play the quiz with kids through algebra to make kids love it.
