Problem 11 - Calculus an Applied Approach

Question

Find \(d y / d x\) \(\frac{x+y}{2 x-y}=1\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{1}{2}\)

Step 1: Simplify the equation

Rewrite the equation as \(2 x - y = x + y\). Then simplify it to get \(y = x/2\)

Step 2: Find the derivative

Use the power rule (which states that the derivative of \(x^n\) is \(n x^{n-1}\)) to find the derivative: \(\frac{d y}{d x} = \frac{1}{2}\)

Step 3: Verify the solution

The derivative \(\frac{dy}{dx}\) of the function \(y = x/2\) is indeed constant and equal to 0.5 as calculated. Hence the solution is correct and final.

Key Concepts

Derivative Calculation

Understanding derivative calculation is crucial in calculus as it provides information about the rate at which a quantity changes. In our exercise, the goal is to find the derivative of y with respect to x, notated as \( \frac{dy}{dx} \). This process begins by recognizing the relationship between x and y given by the equation \( \frac{x+y}{2x-y} = 1 \).

The simplification in Step 1 is vital because it transforms the equation into a form where y can be expressed directly as a function of x, specifically \( y = \frac{x}{2} \). Once the function is singled out, finding its derivative becomes straightforward. This derivative tells us how much y changes for a unit change in x, which is a fundamental concept in calculus and aids in understanding the behavior of functions.

Power Rule Differentiation

When it comes to differentiating polynomials, the power rule is a handy tool. It states that if you have a function \( x^n \), where n is a real number, the derivative of that function is \( n\cdot{x}^{n-1} \).

In our exercise, to calculate \( \frac{dy}{dx} \), we apply the power rule to the function \( y = \frac{x}{2} \), which implicitly means \( y = \frac{1}{2}x^1 \). Since the exponent of x is 1, the derivative is the coefficient \( \frac{1}{2} \) multiplied by 1, resulting in the constant \( \frac{1}{2} \).

Such simplicity helps students quickly identify and apply the correct derivative rule, speeding up their calculations and comprehension, especially when dealing with polynomials.

Verifying Solutions in Calculus

The final step in solving calculus problems often involves verifying the solution. This not only ensures accuracy but also reinforces a student's understanding of the concepts applied. In our case, verification involves confirming that the derivative of \( y = \frac{x}{2} \), which we found to be \( \frac{1}{2} \), is correct.

Verification can be done by applying the derivative to various \(x\) values and seeing if the rates of change align with our calculations. Since our derivative is a constant, we expect no change regardless of the \(x\) values, which can be confirmed by plotting or tabulating values. This step is an excellent practice in calculus, teaching students to be meticulous and to understand the importance of checking their work against the original function.