Last updated on July 21st, 2025
We use the derivative of 5xy, which involves using the product rule, as a way to measure how the function changes in response to a slight change in x or y. Derivatives are useful in various real-life situations, such as calculating profit or loss. We will now discuss the derivative of 5xy in detail.
The derivative of 5xy is commonly represented as d/dx (5xy) or (5xy)'. To find the derivative, we use the product rule since it involves two variables multiplied together.
The function 5xy has a defined derivative, indicating it is differentiable with respect to x or y. The key concepts are mentioned below:
Product Rule: A rule for differentiating products of two functions.
Constant Multiple Rule: A rule that allows us to factor out constants during differentiation.
The derivative of 5xy can be denoted as d/dx (5xy) or (5xy)'. The formula we use to differentiate 5xy with respect to x is: d/dx (5xy) = 5y + 5x(dy/dx) The formula applies to all x and y within the domain of the function.
We can derive the derivative of 5xy using proofs. To show this, we will use the product rule for differentiation. There are several methods we use to prove this, such as:
The step-by-step process is demonstrated below: Using Product Rule To prove the differentiation of 5xy using the product rule, We use the formula: If u = 5x and v = y,
then according to the product rule: d/dx (uv) = u'v + uv' Substitute u = 5x and v = y into the formula: d/dx (5xy) = (5 * 1) * y + 5x * (dy/dx) = 5y + 5x(dy/dx)
Hence, proved.
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. T
o 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 5xy.
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 5xy, we generally use f n(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 y is a constant, the derivative simplifies to 5y, as the derivative of y with respect to x is 0. When x is a constant, the derivative is simply 0, as the derivative of a constant is zero.
Students frequently make mistakes when differentiating functions like 5xy. 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 5xy where y = 3x².
Here, we have f(x) = 5xy and y = 3x². Substituting y into f(x): f(x) = 5x(3x²) = 15x³.
Using the power rule: f'(x) = d/dx (15x³) = 45x².
Thus, the derivative of the specified function is 45x².
We find the derivative of the given function by substituting y = 3x² into 5xy. We then simplify and differentiate using the power rule to obtain the final result.
A company manufactures widgets, and the production output is modeled by the function P(x, y) = 5xy, where x is the number of machines and y is the hours of operation. If x = 10 machines and y = 4x hours, find the rate of change of production with respect to x.
We have P(x, y) = 5xy and y = 4x. Substitute y into P(x, y): P(x) = 5x(4x) = 20x².
Differentiating with respect to x: dP/dx = d/dx (20x²) = 40x. Given x = 10, substitute into the derivative: dP/dx = 40(10) = 400.
Hence, the rate of change of production with respect to x is 400.
We find the rate of change of production by substituting y = 4x into P(x, y) to express it as a function of x.
We then differentiate and evaluate at x = 10 to find the result.
Derive the second derivative of the function f(x, y) = 5xy where y = x² + 1.
First, find the first derivative: f(x) = 5x(x² + 1) = 5x³ + 5x.
The first derivative is: f'(x) = d/dx (5x³ + 5x) = 15x² + 5.
Now find the second derivative: f''(x) = d²/dx² (15x² + 5) = 30x.
Therefore, the second derivative of the function is 30x.
We start by expressing f(x, y) as a function of x using y = x² + 1.
After finding the first derivative, we differentiate again to obtain the second derivative.
Prove: d/dx (5x²y) = 10xy + 5x²(dy/dx).
Let's use the product rule: Consider u = 5x² and v = y. d/dx (uv) = u'v + uv'.
Substitute: d/dx (5x²y) = (10x)y + 5x²(dy/dx). = 10xy + 5x²(dy/dx).
Hence, proved.
In this proof, we apply the product rule to differentiate 5x²y. We express the derivative in terms of x and y, considering dy/dx as necessary.
Solve: d/dx (5xy + 2x).
Differentiate each term separately: d/dx (5xy) = 5y + 5x(dy/dx) and d/dx (2x) = 2.
Combine them: d/dx (5xy + 2x) = 5y + 5x(dy/dx) + 2.
Therefore, d/dx (5xy + 2x) = 5y + 5x(dy/dx) + 2.
We differentiate the function by applying the product rule to 5xy and using the basic rule for the linear term 2x. The derivatives are then combined for 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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