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Last updated on July 17th, 2025

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Derivative of 1/t

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We use the derivative of 1/t, which is -1/t², as a measuring tool for how the function changes in response to a slight change in t. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 1/t in detail.

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What is the Derivative of 1/t?

We now understand the derivative of 1/t. It is commonly represented as d/dt (1/t) or (1/t)', and its value is -1/t². The function 1/t has a clearly defined derivative, indicating it is differentiable within its domain.

 

The key concepts are mentioned below:

 

Reciprocal Function: (1/t is the reciprocal of t).

 

Power Rule: Rule for differentiating 1/t (since it can be expressed as t⁻¹).

 

Negative Exponent: Understanding how to differentiate functions with negative exponents.

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Derivative of 1/t Formula

The derivative of 1/t can be denoted as d/dt (1/t) or (1/t)'. The formula we use to differentiate 1/t is: d/dt (1/t) = -1/t² The formula applies to all t where t ≠ 0.

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Proofs of the Derivative of 1/t

We can derive the derivative of 1/t using proofs. To show this, we will use the rules of differentiation.

 

There are several methods we use to prove this, such as:

 

  1. By First Principle
  2. Using Power Rule
  3. Using Quotient Rule

 

We will now demonstrate that the differentiation of 1/t results in -1/t² using the above-mentioned methods:

 

By First Principle

 

The derivative of 1/t can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 1/t using the first principle, we will consider f(t) = 1/t. Its derivative can be expressed as the following limit.

 

f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1)

 

Given that f(t) = 1/t, we write f(t + h) = 1/(t + h).

 

Substituting these into equation (1), f'(t) = limₕ→₀ [1/(t + h) - 1/t] / h = limₕ→₀ [(t - (t + h)) / (t(t + h))] / h = limₕ→₀ [-h / (t² + th)] / h = limₕ→₀ [-1 / (t² + th)]

 

As h approaches 0, we get: f'(t) = -1/t². Hence, proved.

 

Using Power Rule

 

To prove the differentiation of 1/t using the power rule, We rewrite 1/t as t⁻¹. d/dt (t⁻¹) = -1 * t⁻² Since t⁻² = 1/t², d/dt (1/t) = -1/t².

 

Using Quotient Rule We will now prove the derivative of 1/t

 

using the quotient rule.

 

The step-by-step process is demonstrated below: Here, we use the formula, 1/t = 1 * (t)⁻¹ Given that, u = 1 and v = t Using the quotient rule formula: d/dt [u/v] = (v * u' - u * v') / v² u' = d/dt (1) = 0 (substitute u = 1) v' = d/dt (t) = 1 (substitute v = t)

 

Again, use the quotient rule formula: d/dt (1/t) = (t * 0 - 1 * 1) / t² = -1/t²

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Higher-Order Derivatives of 1/t

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 1/t.

 

For the first derivative of a function, we write f′(t), 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′′(t). Similarly, the third derivative, f′′′(t) is the result of the second derivative and this pattern continues.

 

For the nth Derivative of 1/t, we generally use fⁿ(t) for the nth derivative of a function f(t), which tells us the change in the rate of change (continuing for higher-order derivatives).

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Special Cases:

When t is 0, the derivative is undefined because 1/t is undefined at that point. For positive t, the derivative of 1/t is negative, indicating a decreasing function. For negative t, the derivative of 1/t is also negative, indicating an increasing function.

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Common Mistakes and How to Avoid Them in Derivatives of 1/t

Students frequently make mistakes when differentiating 1/t. 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 power 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 1/t

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They might not remember that 1/t is undefined at the point t = 0. 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 this certain point.

Mistake 3

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Incorrect use of Quotient Rule

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While differentiating functions such as 1/t², students misapply the quotient rule.

 

For example: Incorrect differentiation: d/dt (1/t²) = -2/t³. d/dt (u/v) = (v * u' - u * v') / v² (where u = 1 and v = t²)

 

Applying the quotient rule, d/dt (1/t²) = (t² * 0 - 1 * 2t) / t⁴

 

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 1/t.

 

For example, they incorrectly write d/dt (5/t) = -1/t².

 

Students should check the constants in the terms and ensure they are multiplied properly.

 

For e.g., the correct equation is d/dt (5/t) = -5/t².

Mistake 5

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Not Applying the Power Rule

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Students often forget to use the power rule. This happens when the negative exponent is not considered.

 

For example: Incorrect: d/dt (t⁻¹) = 1/t. To fix this error, students should use the power rule for negative exponents.

 

For example, d/dt (t⁻¹) = -1/t².

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Examples Using the Derivative of 1/t

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Problem 1

Calculate the derivative of (1/t · t²)

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Here, we have f(t) = 1/t · t². Using the product rule, f'(t) = u′v + uv′ In the given equation, u = 1/t and v = t².

 

Let’s differentiate each term, u′ = d/dt (1/t) = -1/t² v′ = d/dt (t²) = 2t

 

Substituting into the given equation, f'(t) = (-1/t²) * t² + (1/t) * 2t

 

Let’s simplify terms to get the final answer, f'(t) = -1 + 2 f'(t) = 1

 

Thus, the derivative of the specified function is 1.

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 chemical reaction is monitored over time. The concentration of a reactant decreases in proportion to 1/t, where t is the time in seconds. Find the rate of change of concentration at t = 5 seconds.

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We have concentration = 1/t

 

Now, we will differentiate the equation Take the derivative 1/t: d/dt (1/t) = -1/t²

 

Given t = 5 seconds (substitute this into the derivative)

 

d/dt (1/5) = -1/5² d/dt (1/5) = -1/25

 

Hence, the rate of change of concentration at t = 5 seconds is -1/25.

Explanation

We find the rate of change of concentration at t = 5 seconds as -1/25, which means that the concentration is decreasing at this rate per second at t = 5 seconds.

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Problem 3

Derive the second derivative of the function y = 1/t.

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The first step is to find the first derivative, dy/dt = -1/t²... (1)

 

Now we will differentiate equation (1) to get the second derivative: d²y/dt² = d/dt [-1/t²]

 

Here we use the power rule, d²y/dt² = 2/t³

 

Therefore, the second derivative of the function y = 1/t is 2/t³.

Explanation

We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate -1/t². We then simplify the terms to find the final answer.

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Problem 4

Prove: d/dt (1/t²) = -2/t³.

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Let’s start using the power rule: Consider y = 1/t² y = t⁻²

 

To differentiate, we use the power rule: dy/dt = -2 * t⁻³

 

Since t⁻³ = 1/t³, dy/dt = -2/t³

 

Substituting y = 1/t², d/dt (1/t²) = -2/t³

 

Hence proved.

Explanation

In this step-by-step process, we used the power rule to differentiate the equation. Then, we replace t⁻² with its derivative. As a final step, we substitute y = 1/t² to derive the equation.

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Problem 5

Solve: d/dt (1/t / t)

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To differentiate the function, we use the quotient rule: d/dt (1/t / t) = (d/dt (1/t) * t - 1/t * d/dt (t)) / t²

 

We will substitute d/dt (1/t) = -1/t² and d/dt (t) = 1 = (-1/t² * t - 1/t * 1) / t² = (-1/t - 1/t) / t² = -2/t³

 

Therefore, d/dt (1/t / t) = -2/t³

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 1/t

1.Find the derivative of 1/t.

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2.Can we use the derivative of 1/t in real life?

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3.Is it possible to take the derivative of 1/t at the point where t = 0?

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4.What rule is used to differentiate 1/t / t?

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5.Are the derivatives of 1/t and ln(t) the same?

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6.Can we find the derivative of the 1/t formula?

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Important Glossaries for the Derivative of 1/t

  • Derivative: The derivative of a function indicates how the given function changes in response to a slight change in t.

 

  • Reciprocal Function: A function expressed as 1 divided by a variable, such as 1/t.

 

  • Power Rule: A basic rule of differentiation applicable to power functions which helps find derivatives of terms like tⁿ.

 

  • Negative Exponent: An exponent that is a negative number, often used in calculus to express reciprocal functions.

 

  • Undefined Point: A point at which a mathematical expression has no meaning, such as 1/t at t = 0.
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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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