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Last updated on July 15th, 2025

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Derivative of Arctan u

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We use the derivative of arctan(u), which is 1/(1 + u²), as a tool to understand how the arctangent function changes in response to variations in u. Derivatives are crucial in fields like economics and physics for calculating profit, loss, and rates of change. We will now discuss the derivative of arctan(u) in detail.

Derivative of Arctan u for US Students
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What is the Derivative of Arctan u?

We understand the derivative of arctan(u). It is commonly represented as d/du (arctan(u)) or (arctan(u))', and its value is 1/(1 + u²). The function arctan(u) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Arctangent Function: (arctan(u) is the inverse of tan(u)). Chain Rule: Rule for differentiating composite functions like arctan(u). Reciprocal Function: A function that returns the reciprocal of its input, used in deriving the formula.

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Derivative of Arctan u Formula

The derivative of arctan(u) can be denoted as d/du (arctan(u)) or (arctan(u))'. The formula we use to differentiate arctan(u) is: d/du (arctan(u)) = 1/(1 + u²) This formula applies to all u where the function is defined.

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Proofs of the Derivative of Arctan u

We can derive the derivative of arctan(u) using proofs. To show this, we will use trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule We will now demonstrate that differentiating arctan(u) results in 1/(1 + u²) using the above-mentioned methods: By First Principle The derivative of arctan(u) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of arctan(u) using the first principle, we will consider f(u) = arctan(u). Its derivative can be expressed as the following limit. f'(u) = limₕ→₀ [f(u + h) - f(u)] / h Given that f(u) = arctan(u), we write f(u + h) = arctan(u + h). Substituting these into the limit equation, we use trigonometric identities to simplify: f'(u) = limₕ→₀ [arctan(u + h) - arctan(u)] / h = limₕ→₀ [tan(arctan(u + h)) - tan(arctan(u))] / [h(1 + u²)] Using the identity tan(arctan(x)) = x, we simplify further to show: f'(u) = 1/(1 + u²) Hence, proved. Using Chain Rule To prove the differentiation of arctan(u) using the chain rule, We use the identity that relates arctan(u) to its derivative: Let y = arctan(u) ⇒ tan(y) = u Differentiating both sides with respect to u, using implicit differentiation: sec²(y) dy/du = 1 Using the identity 1 + tan²(y) = sec²(y), substitute tan(y) = u: (1 + u²) dy/du = 1 Thus, dy/du = 1/(1 + u²) Hence, proved.

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Higher-Order Derivatives of Arctan u

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 arctan(u). For the first derivative of a function, we write f′(u), 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′′(u). Similarly, the third derivative, f′′′(u), is the result of the second derivative, and this pattern continues. For the nth Derivative of arctan(u), we generally use f⁽ⁿ⁾(u) for the nth derivative of a function f(u), which tells us the change in the rate of change, continuing for higher-order derivatives.

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Special Cases:

When u approaches ±∞, the derivative approaches 0 because the slope of arctan(u) flattens out. When u = 0, the derivative of arctan(u) = 1/(1 + 0²), which is 1.

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Common Mistakes and How to Avoid Them in Derivatives of Arctan u

Students frequently make mistakes when differentiating arctan(u). 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 Expression

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Students may forget to simplify the expression, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result, especially when solving using the 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

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Forgetting Undefined Points

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They might not remember that arctan(u) is undefined at certain points outside the domain. 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 Chain Rule

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While differentiating functions involving arctan(u), students misapply the chain rule. For example, incorrect differentiation: d/du (arctan(u²)) = 2u/(1 + u²). Correct differentiation requires applying the chain rule: d/du (arctan(u²)) = 2u/(1 + (u²)²). To avoid this mistake, ensure that the chain rule is applied correctly. 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 where students at times forget to multiply the constants placed before arctan(u). For example, they incorrectly write d/du (5 arctan(u)) = 1/(1 + u²). Students should check the constants in the terms and ensure they are multiplied properly. For e.g., the correct equation is d/du (5 arctan(u)) = 5/(1 + u²).

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/du (arctan(3u)) = 1/(1 + (3u)²). Correct: d/du (arctan(3u)) = 3/(1 + (3u)²). To fix this error, students should divide the functions into inner and outer parts. Then, make sure that each function is differentiated.

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Examples Using the Derivative of Arctan u

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Problem 1

Calculate the derivative of (arctan(u) · u³)

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Here, we have f(u) = arctan(u) · u³. Using the product rule, f'(u) = u′v + uv′ In the given equation, u = arctan(u) and v = u³. Let’s differentiate each term, u′= d/du (arctan(u)) = 1/(1 + u²) v′= d/du (u³) = 3u² Substituting into the given equation, f'(u) = (1/(1 + u²)) · u³ + arctan(u) · 3u² Let’s simplify terms to get the final answer, f'(u) = u³/(1 + u²) + 3u² · arctan(u) Thus, the derivative of the specified function is u³/(1 + u²) + 3u² · arctan(u).

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 bridge's angle of elevation from a certain point on the ground is represented by the function y = arctan(u) where y represents the angle in radians as a function of the horizontal distance u. If u = 1 meter, calculate the rate of change of the angle of elevation.

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We have y = arctan(u) (angle of elevation)...(1) Now, we will differentiate equation (1) Take the derivative arctan(u): dy/du = 1/(1 + u²) Given u = 1 (substitute this into the derivative) dy/du = 1/(1 + 1²) dy/du = 1/2 Hence, the rate of change of the angle of elevation at u = 1 meter is 1/2 radians per meter.

Explanation

We find the rate of change of the angle of elevation at u = 1 meter as 1/2 radians per meter, indicating how the angle changes with respect to the horizontal distance.

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Problem 3

Derive the second derivative of the function y = arctan(u).

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The first step is to find the first derivative, dy/du = 1/(1 + u²)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/du² = d/du [1/(1 + u²)] Here we use the chain rule, d²y/du² = -2u/(1 + u²)² Therefore, the second derivative of the function y = arctan(u) is -2u/(1 + u²)².

Explanation

We use the step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate 1/(1 + u²). We then simplify the terms to find the final answer.

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Problem 4

Prove: d/du (arctan(u²)) = 2u/(1 + u⁴).

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Let’s start using the chain rule: Consider y = arctan(u²) To differentiate, we use the chain rule: dy/du = d/du [arctan(u²)] Using the derivative of arctan(v) where v = u²: dy/du = 1/(1 + (u²)²) · d/du(u²) dy/du = 1/(1 + u⁴) · 2u d/du (arctan(u²)) = 2u/(1 + u⁴) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. We replaced u² with its derivative. As a final step, we simplified the expression to derive the equation.

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Problem 5

Solve: d/du (arctan(u)/u)

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To differentiate the function, we use the quotient rule: d/du (arctan(u)/u) = (d/du (arctan(u)) · u - arctan(u) · d/du(u))/u² We will substitute d/du (arctan(u)) = 1/(1 + u²) and d/du (u) = 1 = (1/(1 + u²) · u - arctan(u) · 1)/u² = (u/(1 + u²) - arctan(u))/u² Therefore, d/du (arctan(u)/u) = (u/(1 + u²) - arctan(u))/u²

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.

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FAQs on the Derivative of Arctan u

1.Find the derivative of arctan(u).

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2.Can we use the derivative of arctan(u) in real life?

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3.Is it possible to take the derivative of arctan(u) at the point where u = ±∞?

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4.What rule is used to differentiate arctan(u)/u?

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5.Are the derivatives of arctan(u) and arctan⁻¹(u) the same?

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6.Can we find the derivative of the arctan(u) formula?

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Important Glossaries for the Derivative of Arctan u

Derivative: The derivative of a function indicates how the given function changes in response to a slight change in its input. Arctangent Function: The arctangent function is the inverse of the tangent function and is written as arctan(u). Chain Rule: A rule for differentiating compositions of functions. First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function. Implicit Differentiation: A method used to find derivatives of functions not explicitly solved for one variable in terms of another.

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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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