Problem 21 - Calculus - AP Edition

Question

$$\text { Find } d y /\left.d x\right|_{x=1}$$. $$y=\frac{2 x-1}{x+3}$$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) evaluated at \( x = 1 \) is \( \frac{7}{16} \).

Step 1: Understand the Problem

We need to find the derivative of the function \(y = \frac{2x - 1}{x + 3}\) evaluated at \(x = 1\). This means determining \( \frac{dy}{dx} \) and then substituting \( x = 1 \) into it.

Step 2: Identify the Derivative Form

The function is a quotient, so we will use the quotient rule for derivatives. The quotient rule states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). In this case, \( u = 2x - 1 \) and \( v = x + 3 \).

Step 3: Differentiate the Numerator and Denominator

Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).- \( u = 2x - 1 \) implies \( \frac{du}{dx} = 2 \).- \( v = x + 3 \) implies \( \frac{dv}{dx} = 1 \).

Step 4: Apply the Quotient Rule

Substitute \( u, v, \frac{du}{dx}, \) and \( \frac{dv}{dx} \) into the quotient rule:\[\frac{dy}{dx} = \frac{(x + 3) \cdot 2 - (2x - 1) \cdot 1}{(x + 3)^2}.\]

Step 5: Simplify the Derivative Expression

Simplify the expression to find \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{2x + 6 - 2x + 1}{(x + 3)^2} = \frac{7}{(x + 3)^2}.\]

Step 6: Evaluate the Derivative at x = 1

Substitute \( x = 1 \) into the derivative \( \frac{dy}{dx} = \frac{7}{(x + 3)^2} \):\[\frac{dy}{dx} \Big|_{x = 1} = \frac{7}{(1 + 3)^2} = \frac{7}{16}.\]

Key Concepts

Quotient Rule

The Quotient Rule is an important tool in calculus for finding the derivative of a function that is the ratio of two differentiable functions. The general formula for this rule is quite straightforward. If you have a function in the form of a fraction \( y = \frac{u}{v} \), where \( u \) and \( v \) are both functions of \( x \), the derivative \( \frac{dy}{dx} \) is given by:

• \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)

This formula may look complex at first, but it's just a way to incorporate both the rate of change of the numerator and denominator, adjusting for the fact that they move together when combined as a fraction.
You essentially first multiply the derivative of the numerator by the denominator and then subtract the product of the numerator and the derivative of the denominator.
Finally, you divide the whole expression by the square of the denominator.

Numerical Differentiation

In calculus, calculating derivatives analytically is a common technique, but what if you want just a numerical approximation? Numerical differentiation might seem a little foreign if you've only worked with algebraic formulas, but it's an essential tool.
Here's how it works: You observe the rate of change over a very small interval. Imagine you're graphing a function and you don't have the exact derivative but you can measure how steep the curve is at small points. This is where methods like finite differences come into play.

• We approximate \( \frac{dy}{dx} \approx \frac{f(x+h) - f(x)}{h} \)

This formula is a forward difference method with small \( h \). It's not perfect but can give you a close estimate of the derivative.

Evaluating Derivatives

After finding the general formula for the derivative of a function, the next step is to evaluate it at specific points to get practical results. Evaluating derivatives might seem straightforward but it's a critical step to verify the understanding of function behavior. For example, after applying the quotient rule and obtaining a derivative expression, you plug in the desired value of \( x \).
This step is about seeing how steep or flat the graph is at a certain point, which is invaluable for understanding critical points and behavior of the function around this area.

• Substitute the value of \( x \) into the derivative expression.
• Solve the resulting expression to find the slope at that particular \( x \).

It's the difference between theoretically knowing the rate of change and seeing it in action at a specific location on the curve.