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103 LearnersLast updated on September 15, 2025
Derivative of Trig Functions Squared
We use the derivatives of trigonometric functions squared to understand how these functions change in response to a slight change in x. Derivatives help us in various real-life applications, such as calculating rates of change. We will now discuss the derivatives of trigonometric functions squared in detail.
What is the Derivative of a Trig Function Squared?
We now explore the derivative of trig functions squared. For example, the derivative of (tan x)² is commonly represented as d/dx ((tan x)²) or ((tan x)²)', and its value is 2 tan x sec²x.
Trigonometric functions like tan x have well-defined derivatives within their domains, indicating they are differentiable. The key concepts are mentioned below:
Tangent Function: (tan(x) = sin(x)/cos(x)).
Square of a Function: When a function is squared, its derivative involves the original function and its derivative.
Secant Function: sec(x) = 1/cos(x).
Derivative of Trig Functions Squared Formula
The derivative of a squared trig function like (tan x)² can be denoted as d/dx ((tan x)²) or ((tan x)²)'.
The formula we use is: d/dx ((tan x)²) = 2 tan x sec²x
This formula applies to all x where the original function is defined and differentiable.
Proofs of the Derivative of Trig Functions Squared
We can derive the derivative of a squared trig function using proofs and rules of differentiation. There are several methods to prove this, such as:
- Using the Chain Rule
- Using the Product Rule
We will now demonstrate the differentiation of (tan x)² using these methods:
Using the Chain Rule
To prove the differentiation of (tan x)² using the chain rule, consider y = (tan x)².
We use the formula: dy/dx = 2 tan x · d/dx (tan x)
Since the derivative of tan x is sec²x, dy/dx = 2 tan x · sec²x
Hence, d/dx ((tan x)²) = 2 tan x sec²x.
Using the Product Rule
We will now prove the derivative of (tan x)² using the product rule.
Here, we consider: (tan x)² = (tan x) · (tan x)
Using the product rule formula: d/dx [u · v] = u' · v + u · v'
Let u = tan x and v = tan x, u' = sec²x and v' = sec²x
d/dx ((tan x)²) = (sec²x) · (tan x) + (tan x) · (sec²x) = 2 tan x sec²x
Thus, the derivative of (tan x)² is 2 tan x sec²x.
Higher-Order Derivatives of Trig Functions Squared
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be complex.
For instance, the first derivative indicates how the function changes, while the second derivative measures the rate of that change. Higher-order derivatives help in understanding functions like (tan x)².
For the first derivative, we write f′(x), which indicates how the function changes at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is the derivative of the second derivative.
For the nth Derivative of (tan x)², we use fⁿ(x) for the nth derivative, indicating the change in the rate of change.
Special Cases:
When x is π/2, the derivative is undefined because tan(x) has a vertical asymptote there. When x is 0, the derivative of (tan x)² = 2 tan(0) sec²(0), which is 0.
Common Mistakes and How to Avoid Them in Derivatives of Trig Functions Squared
Students frequently make mistakes when differentiating trig functions squared. 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 Tan x
They might not remember that tan x is undefined at points such as (x = π/2, 3π/2,...). Keep in mind the domain of the function you differentiate. It will help you understand that the function is not continuous at certain points.
Mistake 3
Incorrect Use of Product Rule
While differentiating functions like (tan x)², students may misapply the product rule. For example: Incorrect differentiation: d/dx ((tan x)²) = sec²x. Applying the product rule correctly: d/dx ((tan x)²) = 2 tan x sec²x 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 where students forget to multiply constants placed before a function. For example, they incorrectly write d/dx (5(tan x)²) = 10 sec²x. Students should check the constants in the terms and ensure they are multiplied properly. For example, the correct equation is d/dx (5(tan x)²) = 10 tan x sec²x.
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 ((tan(2x))²) = 2 tan(2x) sec²(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 ((tan(2x))²) = 4 tan(2x) sec²(2x).
Examples Using the Derivative of Trig Functions Squared
Problem 1
Calculate the derivative of ((sin x)² · sec²x)
Here, we have f(x) = (sin x)² · sec²x.
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = (sin x)² and v = sec²x.
Let’s differentiate each term, u′ = d/dx ((sin x)²) = 2 sin x cos x v′ = d/dx (sec²x) = 2 sec²x tan x
Substituting into the given equation, f'(x) = (2 sin x cos x) · (sec²x) + ((sin x)²) · (2 sec²x tan x)
Let’s simplify terms to get the final answer, f'(x) = 2 sin x cos x sec²x + 2 (sin x)² sec²x tan x
Thus, the derivative of the specified function is 2 sin x cos x sec²x + 2 (sin x)² sec²x tan 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 measures the brightness of a light source using the function y = (cos(x))², where y represents the intensity at angle x. If x = π/6 radians, calculate the rate of change in intensity.
We have y = (cos(x))² (intensity of the light)...(1)
Now, we will differentiate equation (1) Take the derivative of (cos(x))²: dy/dx = 2 cos(x)(-sin(x))
Substitute x = π/6 into the derivative: dy/dx = 2 cos(π/6)(-sin(π/6)) dy/dx = 2(√3/2)(-1/2) dy/dx = -√3/2
Hence, the rate of change in intensity at x = π/6 is -√3/2.
Explanation
We calculate the rate of change of light intensity at x = π/6 by differentiating the squared function and substituting the given angle to find the result.
Problem 3
Derive the second derivative of the function y = (sin(x))².
The first step is to find the first derivative: dy/dx = 2 sin(x) cos(x)...(1)
Now, differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2 sin(x) cos(x)]
Using the product rule, d²y/dx² = 2[cos(x) cos(x) - sin(x) sin(x)] = 2[cos²(x) - sin²(x)]
Therefore, the second derivative of the function y = (sin(x))² is 2[cos²(x) - sin²(x)].
Explanation
We use the step-by-step process to find the first derivative and then apply the product rule to differentiate further, simplifying to get the second derivative.
Problem 4
Prove: d/dx ((cos(x))²) = -2 cos(x) sin(x).
Let’s start using the chain rule: Consider y = (cos(x))² = [cos(x)]²
To differentiate, we use the chain rule: dy/dx = 2 cos(x) · d/dx [cos(x)]
Since the derivative of cos(x) is -sin(x), dy/dx = 2 cos(x) · (-sin(x)) dy/dx = -2 cos(x) sin(x) Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replaced cos(x) with its derivative to derive the equation.
Problem 5
Solve: d/dx ((tan x)²/x)
To differentiate the function, we use the quotient rule:
d/dx ((tan x)²/x) = (d/dx ((tan x)²) · x - (tan x)² · d/dx(x))/x²
Substitute d/dx ((tan x)²) = 2 tan x sec²x and d/dx(x) = 1 = (2 tan x sec²x · x - (tan x)² · 1) / x² = (2 x tan x sec²x - (tan x)²) / x²
Therefore, d/dx ((tan x)²/x) = (2 x tan x sec²x - tan²x) / 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 Trig Functions Squared
1.How do you find the derivative of (tan x)²?
2.Can the derivative of a trig function squared be applied in real life?
3.Can you differentiate (tan x)² at x = π/2?
4.What rule is used to differentiate (tan x)²/x?
5.Are the derivatives of (tan x)² and tan⁻¹x the same?
Important Glossaries for the Derivative of Trig Functions Squared
- Derivative: A derivative of a function indicates how the given function changes in response to a slight change in x.
- Trig Function: A function based on angles, such as sine, cosine, or tangent, and their squares.
- Chain Rule: A rule for finding the derivative of a composite function.
- Product Rule: A rule used to differentiate products of two or more functions.
- Quotient Rule: A rule used to differentiate a function that is the ratio of two differentiable 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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