Last updated on July 22nd, 2025
We use the derivative of xsin(x) to understand 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 talk about the derivative of xsin(x) in detail.
We now understand the derivative of xsin(x). It is commonly represented as d/dx (xsin(x)) or (xsin(x))', and can be found using the product rule. The function xsin(x) is differentiable within its domain, indicating it has a clearly defined derivative.
The key concepts are mentioned below:
Product Rule: Rule for differentiating a product of two functions, such as xsin(x).
Trigonometric Functions: Functions like sin(x), which are fundamental in calculus.
The derivative of xsin(x) can be denoted as d/dx (xsin(x)) or (xsin(x))'. The formula we use to differentiate xsin(x) is: d/dx (xsin(x)) = xcos(x) + sin(x) This formula applies to all x.
We can derive the derivative of xsin(x) using proofs. To show this, we will use the product rule along with the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of xsin(x) results in xcos(x) + sin(x) using the above-mentioned methods:
The derivative of xsin(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of xsin(x) using the first principle, we will consider f(x) = xsin(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) = xsin(x), we write f(x + h) = (x + h)sin(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [(x + h)sin(x + h) - xsin(x)] / h = limₕ→₀ [(xsin(x + h) + hsin(x + h) - xsin(x))] / h = limₕ→₀ [x(sin(x + h) - sin(x)) + hsin(x + h)] / h = limₕ→₀ [x(cos(x)h) + hsin(x + h)] / h = limₕ→₀ [xcos(x) + sin(x + h)]
Using limit formulas, limₕ→₀ sin(x + h) = sin(x). f'(x) = xcos(x) + sin(x).
To prove the differentiation of xsin(x) using the product rule, We use the formula: d/dx [u.v] = u'.v + u.v'
Here, let u = x and v = sin(x). u' = d/dx (x) = 1 v' = d/dx (sin(x)) = cos(x)
By the product rule: d/dx (xsin(x)) = (1)(sin(x)) + (x)(cos(x)) = xcos(x) + sin(x)
We can also use trigonometric identities to simplify the process. Consider xsin(x) as a product of x and sin(x), and differentiate using known identities and rules.
The derivative formula follows from the use of basic trigonometric identity simplifications and standard derivative rules.
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 π/2, the derivative is xcos(x) + sin(x), which simplifies to 0, as cos(π/2) = 0 and sin(π/2) = 1. When x is 0, the derivative of xsin(x) = xcos(x) + sin(x), which simplifies to 0, as both terms become 0.
Students frequently make mistakes when differentiating xsin(x). 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 (xsin(x)·cos(x))
Here, we have f(x) = xsin(x)·cos(x).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = xsin(x) and v = cos(x).
Let’s differentiate each term, u′= d/dx (xsin(x)) = xcos(x) + sin(x) v′= d/dx (cos(x)) = -sin(x)
Substitute into the given equation, f'(x) = (xcos(x) + sin(x))·cos(x) + (xsin(x))·(-sin(x))
Let’s simplify terms to get the final answer, f'(x) = xcos²(x) + sin(x)cos(x) - xsin²(x)
Thus, the derivative of the specified function is xcos²(x) + sin(x)cos(x) - xsin²(x).
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative using the product rule and then combining them to get the final result.
XYZ Construction Company is building a bridge, and the load on the bridge is represented by the function y = xsin(x), where y represents the load at a distance x. If x = π/3 meters, measure the rate of change of the load.
We have y = xsin(x) (load on the bridge)...(1)
Now, we will differentiate the equation (1) Take the derivative of xsin(x): dy/dx = xcos(x) + sin(x)
Given x = π/3 (substitute this into the derivative)
dy/dx = (π/3)cos(π/3) + sin(π/3) = (π/3)(1/2) + (√3/2) = π/6 + √3/2
Hence, we get the rate of change of the load on the bridge at a distance x = π/3 as π/6 + √3/2.
We find the rate of change of the load at x = π/3, which indicates how the load changes at that specific point on the bridge.
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 for xcos(x): d/dx[xcos(x)] = x(-sin(x)) + cos(x) d/dx[sin(x)] = cos(x) So, d²y/dx² = x(-sin(x)) + cos(x) + cos(x) = x(-sin(x)) + 2cos(x)
Therefore, the second derivative of the function y = xsin(x) is x(-sin(x)) + 2cos(x).
We use a step-by-step process to find the first derivative. Using the product rule, we differentiate xcos(x) and then use the derivative of sin(x) to find the final answer.
Prove: d/dx (x²sin(x)) = 2xsin(x) + x²cos(x).
Let’s start using the product rule: Consider y = x²sin(x)
To differentiate, we use the product rule: dy/dx = u'v + uv' Here, let u = x² and v = sin(x). u' = d/dx (x²) = 2x v' = d/dx (sin(x)) = cos(x)
Substitute into the product rule: dy/dx = (2x)(sin(x)) + (x²)(cos(x)) = 2xsin(x) + x²cos(x)
Hence proved.
In this step-by-step process, we used the product rule to differentiate the equation. Then, we replace x² and sin(x) with their derivatives to derive the equation.
Solve: d/dx (xsin(x)/x)
To differentiate the function, we use simplification: d/dx (xsin(x)/x) = d/dx (sin(x)) = cos(x) Therefore, d/dx (xsin(x)/x) = cos(x)
In this process, we simplify the function before differentiating it. The simplification allows us to directly find the derivative of sin(x).
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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