Last updated on July 21st, 2025
We use the derivative of 3sec(x), which is 3sec(x)tan(x), as a tool to understand how the secant function changes with respect to x. Derivatives allow us to calculate rates of change in various real-world scenarios. We will now explore the derivative of 3sec(x) in detail.
The derivative of 3sec(x) is commonly represented as d/dx (3sec(x)) or (3sec(x))', and its value is 3sec(x)tan(x). The function 3sec(x) has a well-defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Secant Function: sec(x) = 1/cos(x).
Product Rule: Used for differentiating a product of functions.
Chain Rule: A method for finding the derivative of composite functions.
The derivative of 3sec(x) can be denoted as d/dx (3sec(x)) or (3sec(x))'. The formula we use to differentiate 3sec(x) is: d/dx (3sec(x)) = 3sec(x)tan(x)
The formula applies to all x where cos(x) ≠ 0
The derivative of 3sec(x) can be derived using proofs. To show this, we use trigonometric identities along with differentiation rules.
There are several methods to prove this, such as:
To prove the differentiation of 3sec(x) using the chain rule, We start with the function sec(x) = 1/cos(x). Consider u(x) = sec(x), then 3u(x) = 3sec(x).
Using the chain rule, the derivative of u(x) is sec(x)tan(x).
Therefore, d/dx [3sec(x)] = 3 * sec(x)tan(x).
We can also prove the derivative of 3sec(x) using the product rule. The step-by-step process is demonstrated below: Let u(x) = 3 and v(x) = sec(x).
Using the product rule: d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Here, u'(x) = 0 and v'(x) = sec(x)tan(x).
Substituting these into the product rule gives: d/dx [3sec(x)] = 3 * sec(x)tan(x).
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be complex, but they help analyze functions in depth.
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 3sec(x), we generally use fⁿ(x) to represent the nth derivative, indicating the change in the rate of change.
When x is π/2, the derivative is undefined because sec(x) has a vertical asymptote there. When x is 0, the derivative of 3sec(x) = 3sec(0)tan(0), which is 0.
Students frequently make mistakes when differentiating 3sec(x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (3sec(x) * tan(x)).
Here, we have f(x) = 3sec(x) * tan(x).
Using the product rule, f'(x) = u′v + uv′. In the given equation, u = 3sec(x) and v = tan(x).
Let’s differentiate each term, u′ = d/dx (3sec(x)) = 3sec(x)tan(x) v′ = d/dx (tan(x)) = sec²(x).
Substituting into the given equation, f'(x) = (3sec(x)tan(x)) * (tan(x)) + (3sec(x)) * (sec²(x)).
Let’s simplify terms to get the final answer, f'(x) = 3sec(x)tan²(x) + 3sec³(x).
Thus, the derivative of the specified function is 3sec(x)tan²(x) + 3sec³(x).
We find the derivative of the given function by dividing it into two parts. First, we find the derivatives of each part and then combine them using the product rule to get the final result.
A company designed a billboard whose height is modeled by the function y = 3sec(x), where y represents the visibility of the billboard at an angle x. If x = π/6 radians, determine the rate of change of visibility.
We have y = 3sec(x) (visibility of the billboard)...(1) Now, we will differentiate equation (1).
Take the derivative of 3sec(x): dy/dx = 3sec(x)tan(x).
Given x = π/6, substitute this into the derivative: dy/dx = 3sec(π/6)tan(π/6).
We know that sec(π/6) = 2/√3 and tan(π/6) = 1/√3. dy/dx = 3 * (2/√3) * (1/√3) = 6/3 = 2.
Hence, the rate of change of visibility of the billboard at x = π/6 is 2.
We find the rate of change of visibility at x = π/6 as 2, which indicates that the visibility changes at a rate of 2 units per radian at that angle.
Derive the second derivative of the function y = 3sec(x).
The first step is to find the first derivative, dy/dx = 3sec(x)tan(x)...(1).
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3sec(x)tan(x)].
Here we use the product rule, d²y/dx² = 3 * [sec(x)tan²(x) + sec³(x)].
Therefore, the second derivative of the function y = 3sec(x) is 3[sec(x)tan²(x) + sec³(x)].
We use a step-by-step process, starting with the first derivative. Using the product rule, we differentiate 3sec(x)tan(x). We then simplify the terms to find the final answer.
Prove: d/dx (9sec²(x)) = 18sec²(x)tan(x).
Let’s start using the chain rule: Consider y = 9sec²(x) = [3sec(x)]².
To differentiate, we use the chain rule: dy/dx = 2 * 3sec(x) * d/dx [3sec(x)].
Since the derivative of 3sec(x) is 3sec(x)tan(x), dy/dx = 2 * 3sec(x) * 3sec(x)tan(x) = 18sec²(x)tan(x).
Hence proved.
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 = 9sec²(x) to derive the equation.
Solve: d/dx (3sec(x)/x).
To differentiate the function, we use the quotient rule: d/dx (3sec(x)/x) = (d/dx (3sec(x)) * x - 3sec(x) * d/dx(x)) / x².
We will substitute d/dx (3sec(x)) = 3sec(x)tan(x) and d/dx(x) = 1. = (3sec(x)tan(x) * x - 3sec(x)) / x² = (3xsec(x)tan(x) - 3sec(x)) / x².
Therefore, d/dx (3sec(x)/x) = (3xsec(x)tan(x) - 3sec(x)) / x².
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.
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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