Math Problem Statement

find the eqution of the line tangent to the graph of y=4lnx at x=3

Solution

To find the equation of the line tangent to the graph of y=4ln(x)y = 4 \ln(x) at x=3x = 3, we follow these steps:

Step 1: Differentiate the function

The derivative of y=4ln(x)y = 4 \ln(x) gives the slope of the tangent line at any point xx. So, we calculate:

dydx=ddx[4ln(x)]=4x\frac{dy}{dx} = \frac{d}{dx} [4 \ln(x)] = \frac{4}{x}

Step 2: Evaluate the slope at x=3x = 3

Now, substitute x=3x = 3 into the derivative to find the slope of the tangent line at this point:

slope=43\text{slope} = \frac{4}{3}

Step 3: Find the point on the curve at x=3x = 3

Substitute x=3x = 3 into the original function y=4ln(x)y = 4 \ln(x) to find the corresponding yy-coordinate:

y=4ln(3)y = 4 \ln(3)

Using the approximation ln(3)1.0986\ln(3) \approx 1.0986:

y4×1.0986=4.3944y \approx 4 \times 1.0986 = 4.3944

Thus, the point of tangency is (3,4.3944)(3, 4.3944).

Step 4: Write the equation of the tangent line

Using the point-slope form of a line equation:

yy1=m(xx1)y - y_1 = m(x - x_1)

where m=43m = \frac{4}{3}, x1=3x_1 = 3, and y1=4.3944y_1 = 4.3944, we get:

y4.3944=43(x3)y - 4.3944 = \frac{4}{3}(x - 3)

Simplifying the equation:

y=43x4+4.3944y = \frac{4}{3}x - 4 + 4.3944

y=43x+0.3944y = \frac{4}{3}x + 0.3944

Thus, the equation of the tangent line is:

y=43x+0.3944y = \frac{4}{3}x + 0.3944

Would you like more details or clarifications on any part of this process?

Related Questions:

  1. How do you find the slope of a curve at any point using derivatives?
  2. What is the general form of the equation of a tangent line?
  3. How does the natural logarithmic function affect the shape of a curve?
  4. Can the derivative of a logarithmic function be extended to include different bases?
  5. What are other methods to approximate logarithmic values without a calculator?

Tip: Understanding derivatives is key to finding the slopes of tangent lines, which represent instantaneous rates of change.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Tangent Line
Logarithmic Functions

Formulas

Derivative of ln(x): d/dx[ln(x)] = 1/x
Point-Slope form of a line: y - y1 = m(x - x1)

Theorems

Derivative of a Logarithmic Function

Suitable Grade Level

Grades 11-12, College Calculus 1