Problem 91 - Calculus and its applications

Question

Find the equation of the line tangent to the graph of \(y=\left(x^{2}-x\right) \ln (6 x)\) at \(x=2\).

Step-by-Step Solution

Answer

The tangent line equation is \( y - 2\ln(12) = (3\ln(12) + 6)(x - 2) \).

Step 1: Differentiate the function

Our function is given as \( y = (x^2 - x) \ln(6x) \). Use the product rule: \( (uv)' = u'v + uv' \). Let \( u = x^2 - x \) and \( v = \ln(6x) \). First, find the derivatives \( u' = 2x - 1 \) and \( v' = \frac{1}{x} \cdot 6 = \frac{6}{x} \). The derivative of the function is \( y' = (2x - 1)\ln(6x) + (x^2 - x)\cdot \frac{6}{x} \). Simplifying, \( y' = (2x - 1)\ln(6x) + 6x - 6 \).

Step 2: Evaluate the derivative at the given point

We need to find \( y'(2) \). Substitute \( x = 2 \) into the derivative: \( y'(2) = (2(2) - 1)\ln(6 \times 2) + 6 \times 2 - 6 \). This becomes \( y'(2) = 3\ln(12) + 12 - 6 \). Compute \( y'(2) = 3\ln(12) + 6 \).

Step 3: Find the point on the curve

Evaluate the original function \( y = (x^2 - x) \ln(6x) \) at \( x = 2 \). Substitute \( x = 2 \) into the equation: \( y = (2^2 - 2) \ln(6 \times 2) = 2 \ln(12) \). This gives the point (2, \( 2\ln(12) \)).

Step 4: Write the equation of the tangent line

The tangent line at a point (\( x_0, y_0 \)) on a curve with slope \( m \) is given by \( y - y_0 = m(x - x_0) \). From the previous steps, \( m = 3\ln(12) + 6 \) and the point is \( (2, 2\ln(12)) \). Substitute these into the tangent line equation: \( y - 2\ln(12) = (3\ln(12) + 6)(x - 2) \). Simplify to find the equation of the tangent line.

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used when differentiating functions that are multiplied together. If you have two functions, say \( u(x) \) and \( v(x) \), their product is differentiated using the formula:

• \( (uv)' = u'v + uv' \)

This means you differentiate \( u \) while keeping \( v \) unchanged, and then add it to the product of \( u \) times the derivative of \( v \).
In the context of our exercise, the function \( y = (x^2 - x) \ln(6x) \) is a product of \( u(x) = x^2 - x \) and \( v(x) = \ln(6x) \). Implementing the product rule helps break down the differentiation process in manageable steps.

Differentiation

Differentiation involves finding the rate at which a function changes at any point, commonly known as the derivative. In practical terms, it allows us to find the slope of a function at any point, which is critical in determining tangent lines.
For the function \( y = (x^2 - x) \ln(6x) \), differentiation is accomplished by using the product rule. First, find the derivatives:

• \( u' = 2x - 1 \)
• \( v' = \frac{6}{x} \)

Apply the product rule:

• \( y' = (2x - 1)\ln(6x) + (x^2 - x)\frac{6}{x} \)
• Simplifying further gives \( y' = (2x - 1)\ln(6x) + 6x - 6 \)

Using these derivatives effectively allows us to solve problems requiring the slope at a point, as seen with calculating \( y'(2) \).

Equation of a Line

The equation of a tangent line is essentially the equation of a straight line that just 'touches' a curve at a given point. When finding such a line, it’s crucial to know two things:

• The slope of the line, provided by the derivative \( y' \).
• The point on the curve where the line is tangent.

The standard linear equation is:

• \( y - y_0 = m(x - x_0) \)

Here, \( m \) is the slope of the tangent, and \((x_0, y_0)\) is the point of tangency.
In our exercise, after calculating \( y'(2) = 3\ln(12) + 6 \) and the point \((2, 2\ln(12))\), it’s straightforward to substitute these into the equation and solve:

• \( y - 2\ln(12) = (3\ln(12) + 6)(x - 2) \)

Working through this equation provides the final form of the tangent line.