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Last updated on September 27, 2025
We use the derivative of work as a measuring tool for how work changes in response to a slight change in the variables involved. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of work in detail.
We now understand the derivative of work. It is commonly represented as dW/dt, where W is work, and its value depends on the context (such as force, distance, or time). The concept of work has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Work: Work is defined as the force applied over a distance (W = F·d).
Force: Force is any interaction that, when unopposed, will change the motion of an object.
Distance: The amount of space between two points, which is part of the work equation.
The derivative of work can be denoted as dW/dt. The formula we use to differentiate work depends on the specific context: dW/dt = F·d' + F'·d
The formula applies to all situations where force and distance are functions of time and are differentiable.
We can derive the derivative of work using proofs. To show this, we will use physical principles along with the rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate the differentiation of work using the above-mentioned methods:
The derivative of work can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of work using the first principle, consider W(t) = F(d(t)). Its derivative can be expressed as the following limit: W'(t) = limₕ→₀ [W(t + h) - W(t)] / h Given that W(t) = F(d(t)), we write W(t + h) = F(d(t + h)). Substituting these into the equation: W'(t) = limₕ→₀ [F(d(t + h)) - F(d(t))] / h = limₕ→₀ [F(d(t) + h) - F(d(t))] / h = F'(d(t))·d'(t) In this context, work is a function of force and distance, and the derivative reflects how work changes with these variables.
To prove the differentiation of work using the chain rule, We use the formula: W = F(d) Consider F = F(t) and d = d(t), By chain rule: dW/dt = F'(t)·d(t) + F(t)·d'(t) This shows how the rate of change of work depends on the rate of change of force and distance.
We will now prove the derivative of work using the product rule. The step-by-step process is demonstrated below: Here, we use the formula: W = F·d Given that F = F(t) and d = d(t), Using the product rule formula: d/dt [F·d] = F'·d + F·d' This highlights how work's rate of change is affected by both force and distance over time.
When work 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 work.
For the first derivative of work, we write W′(t), which indicates how work changes over time. The second derivative is derived from the first derivative, which is denoted using W′′(t). Similarly, the third derivative, W′′′(t), is the result of the second derivative, and this pattern continues.
For the nth Derivative of work, we generally use W^(n)(t) for the nth derivative of a function W(t), which tells us the change in the rate of change, continuing for higher-order derivatives.
When force is constant, the derivative of work simplifies to dW/dt = F·d'.
When distance is constant, the derivative of work simplifies to dW/dt = F'·d.
When both force and distance change with time, use the full derivative formula.
Students frequently make mistakes when differentiating work. 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 W = (F·d²).
Here, we have W(t) = F(t)·d(t)². Using the product rule, W'(t) = F'(t)·d(t)² + F(t)·2d(t)·d'(t) In the given equation, F = F(t) and d = d(t). Let’s differentiate each term: F' = dF/dt d' = dd/dt Substituting into the given equation, W'(t) = F'(t)·d(t)² + 2F(t)·d(t)·d'(t) Thus, the derivative of the specified function is F'(t)·d(t)² + 2F(t)·d(t)·d'(t).
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.
A construction project involves lifting materials with force F(t) = 5t and distance d(t) = t². Calculate the rate of work done when t = 2 seconds.
We have W(t) = F(t)·d(t) = 5t·t² = 5t³. Now, we will differentiate the equation: W'(t) = d/dt (5t³) = 15t² Given t = 2 seconds, W'(2) = 15(2)² = 60 Hence, the rate of work done at t = 2 seconds is 60 units.
We find the rate of work done at t = 2 seconds by differentiating the work function and evaluating it at the given time.
Derive the second derivative of the function W = F·d.
The first step is to find the first derivative, W'(t) = F'(t)·d(t) + F(t)·d'(t) Now we will differentiate again to get the second derivative: W''(t) = d/dt [F'(t)·d(t) + F(t)·d'(t)] Applying the product rule to each term, W''(t) = F''(t)·d(t) + F'(t)·d'(t) + F'(t)·d'(t) + F(t)·d''(t) = F''(t)·d(t) + 2F'(t)·d'(t) + F(t)·d''(t) Therefore, the second derivative of W = F·d is F''(t)·d(t) + 2F'(t)·d'(t) + F(t)·d''(t).
We use a step-by-step process, starting with the first derivative.
Using the product rule, we differentiate each term to find the second derivative.
Prove: d/dt (F²·d) = 2F·F'·d + F²·d'.
Let’s start using the product rule: Consider W = F²·d To differentiate, we use the product rule: dW/dt = d/dt (F²)·d + F²·d' Using the chain rule for F², d/dt (F²) = 2F·F' Substituting, dW/dt = 2F·F'·d + F²·d' Hence proved.
In this step-by-step process, we used the product rule to differentiate the equation.
Then, we replace F² with its derivative using the chain rule.
As a final step, we substitute and simplify the equation.
Solve: d/dt (F·d/t).
To differentiate the function, we use the quotient rule: d/dt (F·d/t) = (d/dt (F·d)·t - F·d·d/dt (t)) / t² We will substitute d/dt (F·d) = F'·d + F·d' and d/dt (t) = 1 = ((F'·d + F·d')·t - F·d) / t² = (t(F'·d + F·d') - F·d) / t² Therefore, d/dt (F·d/t) = (tF'·d + tF·d' - F·d) / t²
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.
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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