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Last updated on July 19th, 2025

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Derivative of Sin/Cos

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We use the derivative of sin(x)/cos(x), which is tan(x) sec(x), as a measuring tool for how the sine and cosine functions change 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 sin(x)/cos(x) in detail.

Derivative of Sin/Cos for US Students
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What is the Derivative of Sin/Cos?

We now understand the derivative of sin x/cos x. It is commonly represented as d/dx (sin x/cos x) or (sin x/cos x)', and its value is tan(x) sec(x). The function sin x/cos x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: - Sine and Cosine Functions: sin(x) and cos(x). - Quotient Rule: Rule for differentiating sin(x)/cos(x). - Tangent and Secant Functions: tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x).

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Derivative of Sin/Cos Formula

The derivative of sin x/cos x can be denoted as d/dx (sin x/cos x) or (sin x/cos x)'. The formula we use to differentiate sin x/cos x is: d/dx (sin x/cos x) = tan(x) sec(x) The formula applies to all x where cos(x) ≠ 0.

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Proofs of the Derivative of Sin/Cos

We can derive the derivative of sin x/cos x 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: - By First Principle - Using Quotient Rule - Using Product Rule We will now demonstrate that the differentiation of sin x/cos x results in tan(x) sec(x) using the above-mentioned methods: By First Principle The derivative of sin x/cos x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of sin x/cos x using the first principle, we will consider f(x) = sin x/cos 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) = sin x/cos x, we write f(x + h) = sin(x + h)/cos(x + h). Substituting these into equation (1), f'(x) = limₕ→0 [sin(x + h)/cos(x + h) - sin x/cos x] / h = limₕ→0 [(sin(x + h) cos x - sin x cos(x + h)) / (cos x cos(x + h))] / h We now use the formula sin A cos B - sin B cos A = sin(A - B). f'(x) = limₕ→0 [sin h] / [h cos x cos(x + h)] = limₕ→0 [(sin h)/h] * limₕ→0 [1 / (cos x cos(x + h))] Using limit formulas, limₕ→0 (sin h)/h = 1. f'(x) = 1 * [1 / (cos² x)] = 1/cos² x As the reciprocal of cosine is secant, we have, f'(x) = sec² x. Hence, proved. Using Quotient Rule To prove the differentiation of sin x/cos x using the quotient rule, We use the formula: sin x/cos x = tan x Consider f(x) = sin x and g(x) = cos x So, we get tan 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) = sin x and g(x) = cos x in equation (1), d/dx (sin x/cos x) = [(cos x)(cos x) - (sin x)(-sin x)] / (cos x)² = (cos² x + sin² x) / cos² x ... (2) Here, we use the formula: cos² x + sin² x = 1 (Pythagorean identity) Substituting this into (2), d/dx (tan x) = 1 / cos² x Since sec x = 1/cos x, we write: d/dx(tan x) = sec² x Using Product Rule We will now prove the derivative of sin x/cos x using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, sin x/cos x = (sin x)(cos x)⁻¹ Given that u = sin x and v = (cos x)⁻¹ Using the product rule formula: d/dx [u.v] = u'.v + u.v' u' = d/dx (sin x) = cos x (substitute u = sin x) Here we use the chain rule: v = (cos x)⁻¹ (substitute v = (cos x)⁻¹) v' = -1 (cos x)⁻² d/dx (cos x) v' = sin x / cos² x Again, using the product rule formula: d/dx (sin x/cos x) = u'.v + u.v' Let’s substitute u = sin x, u' = cos x, v = (cos x)⁻¹, and v' = sin x / cos² x When we simplify each term: We get, d/dx (sin x/cos x) = 1 + sin²x / cos² x sin² x / cos² x = tan² x (we use the identity sin² x + cos² x = 1) Thus: d/dx (sin x/cos x) = 1 + tan² x Since 1 + tan² x = sec² x d/dx (sin x/cos x) = sec² x.

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Higher-Order Derivatives of Sin/Cos

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 sin(x)/cos(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 sin(x)/cos(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 is π/2, the derivative is undefined because sin(x)/cos(x) has a vertical asymptote there. When x is 0, the derivative of sin(x)/cos(x) = sec²(0), which is 1.

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Common Mistakes and How to Avoid Them in Derivatives of Sin/Cos

Students frequently make mistakes when differentiating sin x/cos 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 Sin/Cos

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They might not remember that sin x/cos 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

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Incorrect use of Quotient Rule

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While differentiating functions like sin x/cos x, students misapply the quotient rule. For example: Incorrect differentiation: d/dx (sin x/cos x) = tan(x) sec(x)/cos²x. d/dx (u/v) = (v . u' - u . v')/v² (where u = sin x and v = cos x) Applying the quotient rule correctly, d/dx (sin x/cos x) = (cos x . d/dx (sin x) - sin x . d/dx (cos x))/cos² x 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 sin x/cos x. For example, they incorrectly write d/dx (5 sin x/cos x) = sec² x. Students should check the constants in the terms and ensure they are multiplied properly. For example, the correct equation is d/dx (5 sin x/cos x) = 5 sec² 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 (sin(2x)/cos(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 (sin(2x)/cos(2x)) = 2 sec²(2x).

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Examples Using the Derivative of Sin/Cos

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Problem 1

Calculate the derivative of (sin x · sec² x)

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Here, we have f(x) = sin x · sec²x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin x and v = sec² x. Let’s differentiate each term, u′= d/dx (sin x) = cos x v′= d/dx (sec² x) = 2 sec² x tan x Substituting into the given equation, f'(x) = (cos x) (sec² x) + (sin x) (2 sec² x tan x) Let’s simplify terms to get the final answer, f'(x) = cos x sec² x + 2 sin x sec² x tan x Thus, the derivative of the specified function is cos x sec² x + 2 sin x sec² x 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.

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Problem 2

A company's profit is modeled by the function y = sin(x)/cos(x) where y represents the profit at time x. If x = π/6 hours, determine the rate of change of profit at this time.

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We have y = sin(x)/cos(x) (rate of change of profit)...(1) Now, we will differentiate the equation (1) Take the derivative sin(x)/cos(x): dy/dx = sec²(x) We know that sec²(x) = 1 + tan²(x) Given x = π/6 (substitute this into the derivative) sec²(π/6) = 1 + tan²(π/6) sec²(π/6) = 1 + (1/√3)² = 4/3 Hence, we get the rate of change of profit at time x = π/6 as 4/3.

Explanation

We find the rate of change of profit at x = π/6 as 4/3, which indicates how profit changes at this specific time.

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Problem 3

Derive the second derivative of the function y = sin(x)/cos(x).

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The first step is to find the first derivative, dy/dx = sec²(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [sec²(x)] Here we use the product rule, d²y/dx² = 2 sec(x) d/dx [sec(x)] d²y/dx² = 2 sec(x) [sec(x) tan(x)] = 2 sec²(x) tan(x) Therefore, the second derivative of the function y = sin(x)/cos(x) is 2 sec²(x) tan(x).

Explanation

We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate sec²(x). We then substitute the identity and simplify the terms to find the final answer.

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Problem 4

Prove: d/dx (sin²(x)/cos²(x)) = 2 tan(x) sec²(x).

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Let’s start using the chain rule: Consider y = sin²(x)/cos²(x) = [tan(x)]² To differentiate, we use the chain rule: dy/dx = 2 tan(x) d/dx [tan(x)] Since the derivative of tan(x) is sec²(x), dy/dx = 2 tan(x) sec²(x) Substituting y = tan²(x), d/dx (tan²(x)) = 2 tan(x) sec²(x) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace tan(x) with its derivative. As a final step, we substitute y = tan²(x) to derive the equation.

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Problem 5

Solve: d/dx (sin(x)/x)

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To differentiate the function, we use the quotient rule: d/dx (sin(x)/x) = (d/dx (sin x) · x - sin x · d/dx(x))/x² We will substitute d/dx (sin x) = cos x and d/dx (x) = 1 = (cos x · x - sin x · 1)/x² = (x cos x - sin x)/x² Therefore, d/dx (sin x/x) = (x cos x - sin 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.

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FAQs on the Derivative of Sin/Cos

1.Find the derivative of sin x/cos x.

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2.Can we use the derivative of sin x/cos x in real life?

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

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

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5.Are the derivatives of sin x/cos x and tan⁻¹x the same?

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6.Can we find the derivative of the sin x/cos x formula?

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Important Glossaries for the Derivative of Sin/Cos

- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. - Sine Function: One of the primary trigonometric functions, denoted as sin x. - Cosine Function: Another primary trigonometric function, denoted as cos x. - Tangent Function: The tangent function is derived from sine and cosine and is written as tan x = sin x/cos x. - Quotient Rule: A rule used to differentiate functions that are expressed as a quotient of two other 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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