Problem 12 - Calculus an Applied Approach

Question

Find \(d y / d x\) \(\frac{2 x+y}{x-5 y}=1\)

Step-by-Step Solution

Answer

The derivative of \(y\) with respect to \(x\), \(d y / d x\) given the equation is \(-1/6\).

Step 1: Rewrite the equation

First, rewrite the equation to make the implicit differentiation step easier. The given equation is \(\frac{2 x+y}{x-5 y}=1\). By cross-multiplying, it can be rewritten as \(2x + y = x - 5y\). Collect like terms by subtracting x from both sides gives \(x + 6y = 0\).

Step 2: Apply implicit differentiation

Now, implicit differentiation will be applied. In general, the derivative of \(y\) with respect to \(x\) can be represented as \(d y / d x\) or \(y'\). For \(x\), the derivative will be just \(1\), since the derivative of \(x\) with respect to \(x\) is \(1\). For \(6y\), apply the rule of derivatives that the derivative of a constant times a function is the constant times the derivative of the function. Here, the derivative of \(y\) with respect to \(x\) is \(d y / d x\) or \(y'\). Thus, the derivative of \(6y\) is \(6(d y / d x)\) or \(6y'\). Differentiating the given equation gives: \(1 + 6y' = 0\).

Step 3: Solve for \(d y / d x\)

Finally, solve the equation from Step 2 for the derivative \(y'\), which stands for \(d y / d x\). To isolate \(y'\) subtract \(1\) from both sides, giving \(6y' = -1\). Then divide both sides by \(6\), yielding \(y' = -1/6\). This is the derivative of the function and thus, the solution to the exercise.

Key Concepts

Derivative

A derivative represents the rate at which one quantity changes relative to another. In simpler terms, it tells you how fast or slow a function is increasing or decreasing.
The concept of a derivative is fundamental in calculus, as it helps understand the behavior of functions. In the exercise, we need to find the derivative \(\frac{dy}{dx}\), which denotes how \(y\) changes with respect to \(x\).

• To find the derivative, we use the implicit differentiation process, treating \(y\) as a function of \(x\).
• This is crucial when the relationship between variables isn’t explicit, such as directly solving for \(y\).
• In the solution, after differentiating the equation, \(y'\) is isolated to find that the rate of change or derivative with respect to \(x\) is \(-1/6\).

Implicit Function

Implicit functions are equations where the dependent variable isn't isolated on one side. Unlike explicit functions where \(y = f(x)\), in implicit functions, \(x\) and \(y\) are mixed together.
This means you cannot easily express one in terms of the other.

• An example is \(2x + y = x - 5y\), which appears in the exercise.
• With implicit differentiation, you can find derivatives even when it’s not possible to explicitly solve for one variable.
• This technique involves differentiating both sides of the equation with respect to \(x\), applying the chain rule where required.
• The exercise demonstrates implicit differentiation, which isolates \(y'\) to solve for the derivative \(\frac{dy}{dx}\).

Mathematics Problem Solving

Solving mathematics problems, especially those involving calculus concepts, requires a structured approach. Identifying the type of problem and the methods to apply is crucial.

• Start by carefully rewriting the equation in a manageable form, as seen when \(\frac{2x + y}{x - 5y} = 1\) was rearranged to \(x + 6y = 0\).
• Problems often require multiple steps, such as cross-multiplying or isolating like terms, to simplify them.
• Understanding which rules to apply, like the chain rule for derivatives, helps solve the problem accurately.
• The goal is to logically progress towards finding a solution, such as deriving \(y' = -1/6\) by applying implicit differentiation correctly.

Calculus

Calculus is a field of mathematics that studies change. It’s divided into two main branches: differential and integral calculus. The exercise focuses on differential calculus, which deals with derivatives.

• Differential calculus helps find the slope of a curve at any point by calculating its derivative.
• In the context of the exercise, we used calculus principles to perform implicit differentiation on an equation.
• This method allows for the calculation of the derivative, \(\frac{dy}{dx}\), even when explicit forms of \(y\) are unavailable.
• By understanding calculus, we can solve complex problems involving rates of change and motion.
• These principles are essential for studying any function's behavior and are widely applicable across sciences and engineering.