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Chapter 3: Problem 76

Find the equation of the line tangent to the graph of \(y=e^{3 x} \cdot \ln (4 x)\) at \(x=1\)

Short Answer

Expert verified
The equation of the tangent line is \[ y - e^3 \text{ln}(4) = (3e^3 \text{ln}(4) + e^3)(x - 1). \]

Step by step solution

01

Evaluate the function at x=1

Begin by finding the value of the function at the point where the tangent is to be determined. For the function \[ y = e^{3x} \times \text{ln}(4x), \] substitute \( x = 1 \).\[ y(1) = e^{3 \times 1} \times \text{ln}(4 \times 1) = e^3 \times \text{ln}(4). \]
02

Find the derivative of y

Apply the product rule to differentiate the function. The product rule states that if \( y = u \times v \), then \( y' = u' v + uv' \). Let \( u = e^{3x} \) and \( v = \text{ln}(4x) \).u' = 3e^{3x}, v' = \left( \frac{1}{4x} \cdot 4 \right) = \frac{1}{x}.Therefore, the derivative is: \[ y' = (3e^{3x})(\text{ln}(4x)) + e^{3x} \left( \frac{1}{x} \right). \]
03

Evaluate the derivative at x=1

Now, substitute \( x = 1 \) into the derivative to find the slope of the tangent line. \[ y'(1) = (3e^3)(\text{ln}(4)) + e^3 \left( \frac{1}{1} \right) = 3e^3 \times \text{ln}(4) + e^3. \]
04

Form the equation of the tangent line

Using the point-slope form of the equation of a line: \( y - y_1 = m (x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. Here, \( x_1 = 1 \), \( y_1 = e^3 \text{ln}(4) \), and \( m = 3e^3 \text{ln}(4) + e^3 \). Substitute these values into the point-slope form:\[ y - e^3 \text{ln}(4) = (3e^3 \text{ln}(4) + e^3)(x - 1). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which represents the rate at which the function's value changes. The derivative of a function at a point gives the slope of the tangent line to the graph of the function at that point.

To differentiate a function:
  • Identify the function and variable you need to differentiate.
  • Apply the rules of differentiation, such as the power rule, product rule, chain rule, etc.
  • Simplify the result.

For example, when differentiating the function \(y=e^{3x} \times \ln(4x)\), you need to use specific rules tailored to its components.
Product Rule
The product rule is a key method in differentiation used when you need to differentiate a product of two functions. It states that if you have two differentiable functions, \( u \) and \( v \), and their product \( y = u \times v \), then the derivative \( y' \) is given by:
  • \( y' = u' v + uv' \)


In the problem, we have \( y = e^{3x} \times \ln(4x) \). Let \( u = e^{3x} \) and \( v = \ln(4x) \).
By applying the product rule:
  • The derivative of \( u \), noted as \( u' \), is \(3e^{3x}\).
  • The derivative of \( v \), noted as \(v'\), is \(\frac{1}{x}\).

So, the derivative of the product is:
\[ y' = (3e^{3x}) (\ln(4x)) + e^{3x} \left(\frac{1}{x}\right) \]
Point-Slope Form
The point-slope form is used to write the equation of a line when you have its slope and a point on the line. The general format is:
  • \( y - y_1 = m(x - x_1) \)

Here, \( m \) is the slope, and \( (x_1, y_1) \) is a point through which the line passes.

Based on the problem:
  • We found the slope at \( x = 1 \) to be \( 3e^3 \times \ln(4) + e^3 \).
  • The point on the curve is \( (1, e^3 \times \ln(4)) \).

Substituting these values into the point-slope form results in:
\[ y - e^3 \times \ln(4) = (3e^3 \times \ln(4) + e^3)(x - 1) \]

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