Last updated on September 6, 2025
We explore the derivative of xsin(x), which measures how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of xsin(x) in detail.
We now understand the derivative of xsinx.
It is commonly represented as d/dx(xsinx) or (xsinx)', and its value is xcosx + sinx.
The function xsinx has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Product Rule: Rule for differentiating xsinx (since it is a product of x and sinx).
Sine Function: sin(x) is a basic trigonometric function.
Cosine Function: cos(x) is another basic trigonometric function.
The derivative of xsinx can be denoted as d/dx(xsinx) or (xsinx)'.
The formula we use to differentiate xsinx is: d/dx(xsinx) = xcosx + sinx
The formula applies to all x for which the functions involved are defined.
We can derive the derivative of xsinx using proofs.
To show this, we will use the rules of differentiation and trigonometric identities.
There are several methods we use to prove this, such as:
Using the Product Rule
Using the First Principle
Using the Chain Rule
We will now demonstrate that the differentiation of xsinx results in xcosx + sinx using the above-mentioned methods:
Using the Product Rule
To prove the differentiation of xsinx using the product rule, We use the formula: xsinx = (x)(sinx)
Using the product rule formula: d/dx [u·v] = u'·v + u·v'
Let u = x and v = sinx u' = d/dx(x) = 1 v' = d/dx(sinx) = cosx
Substituting into the formula, d/dx(xsinx) = (1)(sinx) + (x)(cosx) = sinx + xcosx
Using the First Principle
The derivative of xsinx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of xsinx using the first principle, we will consider f(x) = xsinx.
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = xsinx, we write f(x + h) = (x + h)sin(x + h).
Substituting these into the equation: f'(x) = limₕ→₀ [(x + h)sin(x + h) - xsinx] / h = limₕ→₀ [xsin(x + h) + hsin(x + h) - xsinx] / h = limₕ→₀ [x(sin(x + h) - sinx) + hsin(x + h)] / h = limₕ→₀ [x(cosx·h) + hsin(x + h)] / h = limₕ→₀ [xcosx + sin(x + h)]
Using limit formulas, limₕ→₀ sin(x + h) = sinx. f'(x) = xcosx + sinx
Hence, proved.
Using the Chain Rule
To prove the differentiation of xsinx using the chain rule, We consider f(x) = xsinx
Let u = sinx, then f(x) = x·u Using the product rule: d/dx(x·u) = x'·u + x·u'
Substitute u = sinx and u' = cosx, d/dx(xsinx) = (1)(sinx) + (x)(cosx) = xcosx + sinx
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 xsin(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 xsin(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).
When x is 0, the derivative of xsinx = xcos(0) + sin(0), which is 0. When x is π/2, the derivative of xsinx = xcos(π/2) + sin(π/2), which is 1.
Students frequently make mistakes when differentiating xsinx. 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 (xsinx·cosx)
Here, we have f(x) = xsinx·cosx.
Using the product rule, f'(x) = u′v + uv′
In the given equation, u = xsinx and v = cosx.
Let’s differentiate each term, u′ = d/dx(xsinx) = xcosx + sinx v′ = d/dx(cosx) = -sinx
Substituting into the given equation, f'(x) = (xcosx + sinx)·(cosx) + (xsinx)·(-sinx)
Let’s simplify terms to get the final answer, f'(x) = xcos²x + sinxcosx - xsin²x
Thus, the derivative of the specified function is xcos²x + sinxcosx - xsin²x.
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.
XYZ Corporation monitors the oscillations of a pendulum. The displacement is represented by the function y = xsin(x) where y represents the displacement at a time x. If x = π/6 seconds, calculate the rate of change of displacement.
We have y = xsin(x) (displacement function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of xsin(x): dy/dx = xcos(x) + sin(x)
Given x = π/6 (substitute this into the derivative) dy/dx = (π/6)cos(π/6) + sin(π/6)
We know that cos(π/6) = √3/2 and sin(π/6) = 1/2. dy/dx = (π/6)(√3/2) + (1/2) dy/dx = (π√3/12) + (1/2)
Hence, we get the rate of change of displacement at time x = π/6 as (π√3/12) + (1/2).
We find the rate of change of displacement at x = π/6 by differentiating the displacement function and substituting the given value of x into the derivative.
Derive the second derivative of the function y = xsin(x).
The first step is to find the first derivative, dy/dx = xcos(x) + sin(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx[xcos(x) + sin(x)] d²y/dx² = d/dx(xcos(x)) + d/dx(sin(x))
Using the product rule on xcos(x), d/dx(xcos(x)) = x·(-sin(x)) + cos(x) = -xsin(x) + cos(x) And d/dx(sin(x)) = cos(x) d²y/dx² = -xsin(x) + cos(x) + cos(x) d²y/dx² = -xsin(x) + 2cos(x)
Therefore, the second derivative of the function y = xsin(x) is -xsin(x) + 2cos(x).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate xcos(x). We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx((xsinx)²) = 2xsinx(xcosx + sinx).
Let’s start using the chain rule:
Consider y = (xsinx)² To differentiate, we use the chain rule: dy/dx = 2(xsinx)·d/dx(xsinx)
Since the derivative of xsinx is xcosx + sinx, dy/dx = 2(xsinx)·(xcosx + sinx)
Substituting y = (xsinx)², d/dx((xsinx)²) = 2xsinx(xcosx + sinx) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace xsinx with its derivative. As a final step, we substitute y = (xsinx)² to derive the equation.
Solve: d/dx(xsinx/x)
To differentiate the function, we use the quotient rule: d/dx(xsinx/x) = (d/dx(xsinx)·x - xsinx·d/dx(x))/x²
We will substitute d/dx(xsinx) = xcosx + sinx and d/dx(x) = 1 = ((xcosx + sinx)·x - xsinx·1) / x² = (x²cosx + xsinx - xsinx) / x² = x²cosx/x² = cosx
Therefore, d/dx(xsinx/x) = cosx
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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