Problem 60 - Calculus and its applications

Question

Find an equation for the tangent line to the graph of \(y=\left(\frac{2 x+3}{x-1}\right)^{3}\) at the point (2,343)

Step-by-Step Solution

Answer

The tangent line equation is \(y = -735x + 1813\).

Step 1: Understand the Function

You're given a function \( y = \left(\frac{2x+3}{x-1}\right)^3 \) and asked to find the equation of the tangent line at the point \((2, 343)\). The tangent line represents the instantaneous rate of change of the function at \( x = 2 \).

Step 2: Differentiate the Function

To find the slope of the tangent line, which is the derivative, utilize the chain rule. Let \( u = \left(\frac{2x+3}{x-1}\right) \), then \( y = u^3 \). Differentiate to get \( \frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx} \).

Step 3: Differentiate the Inner Function

Differentiate \( u = \frac{2x+3}{x-1} \) using the quotient rule. Set \( v = 2x+3 \) and \( w = x-1 \). Using \( \frac{du}{dx} = \frac{wv' - vw'}{w^2} \), calculate the derivatives: \( v' = 2 \) and \( w'= 1 \). Substitute: \( \frac{du}{dx} = \frac{(x-1)(2) - (2x+3)(1)}{(x-1)^2} \).

Step 4: Simplify the Derivative

Simplify to \( \frac{du}{dx} = \frac{2x-2 - (2x+3)}{(x-1)^2} = \frac{-5}{(x-1)^2} \). Substitute \( \frac{du}{dx} \) into the derivative of \( y \): \( \frac{dy}{dx} = 3u^2 \left(\frac{-5}{(x-1)^2}\right) \).

Step 5: Evaluate at the Given Point

At \( x = 2 \), calculate \( u = \frac{2(2)+3}{2-1} = 7 \). Hence, \( y' = \frac{dy}{dx} = 3(7)^2 \left(\frac{-5}{1^2}\right) = -735 \). This is the slope at \( x = 2 \).

Step 6: Form the Equation of the Tangent Line

Use the point-slope form \( y - y_1 = m(x - x_1) \) where \( m = -735 \), \( x_1 = 2 \), \( y_1 = 343 \). Substitute: \( y - 343 = -735(x - 2) \). Simplify to get the equation of the tangent line.

Step 7: Simplify the Tangent Line Equation

The equation simplifies to \( y - 343 = -735x + 1470 \). Therefore, \( y = -735x + 1813 \).

Key Concepts

Derivative

A derivative measures how a function changes as its input changes. It gives the slope of the tangent line to the graph of the function at any given point. This concept is crucial when finding equations of tangent lines, as the slope is an essential part of these lines.
When you're working with the function given in the exercise, the derivative is calculated to find the slope at the specific point where the tangent line touches the curve. In calculus, understanding derivatives allows us to predict how changes in one variable affect another.
In this specific problem, you find the derivative of the function using the chain and quotient rules. This shows the composite nature of the problem, where layers of functions interact. This requires applying these rules to manage each layer of the derivative process correctly.

Point-Slope Form

The point-slope form of a linear equation provides a method to write the equation of a line if you know the slope and a point on the line. It is expressed as:

• \( y - y_1 = m(x - x_1) \)

This formula highlights the relationship between the slope \( m \), and a given point \((x_1, y_1)\). In the context of our exercise, point-slope form lets us transition from the derivative's slope to the actual equation of the tangent line.
In applying this to our problem, we plug in the derivative value calculated at the specific point, known as \( x = 2 \), and the point provided, \((2, 343)\). Allowing a direct way to assemble the equation of the tangent line in a coherent and structured process.

Chain Rule

The chain rule is pivotal when dealing with composite functions like the one in this exercise. It helps us differentiate functions nested within other functions. The chain rule states that you take the derivative of the outer function and multiply it by the derivative of the inner function.
For the function in question, defined as \( y = (\frac{2x+3}{x-1})^3 \), we first identify the inner and outer functions. Here, \( u = \frac{2x+3}{x-1} \) and \( y = u^3 \). To apply the chain rule:

• Differentiating \( y = u^3 \) gives \( \, \frac{dy}{du} = 3u^2 \).
• We then find \( \frac{du}{dx} \) using the next important technique—the quotient rule.

Everything comes together when you apply the chain rule to find the overall derivative \( \frac{dy}{dx} \). This shows how changes in \( x \) affect \( y \), crucial for identifying the slope of the tangent.

Quotient Rule

When a function is expressed as a quotient of two other functions, the quotient rule is necessary to find its derivative. The quotient rule is applied as:

• \( \frac{d}{dx} \left( \frac{v}{w} \right) = \frac{w \frac{d}{dx}(v) - v \frac{d}{dx}(w)}{w^2} \)

This rule ensures accurate derivates of functions divided by each other. For our exercise, \( v = 2x + 3 \) and \( w = x - 1 \).
Following these steps, we find:

• \( v' = 2 \) and \( w' = 1 \)
• Substitute these into \( \frac{du}{dx} = \frac{(x-1)(2) - (2x+3)(1)}{(x-1)^2} \)

Simplifying gives \( \frac{du}{dx} = \frac{-5}{(x-1)^2} \).
This value merges with the outer derivative's chain rule operation, making the equation's differentiation complete and usable for further calculation in constructing the tangent equation.