**Problem Statement:**Find the equation of the tangent line of \( f(x) = x^2 \ln x \) at \( x = 1 \).**Explanation:**To find the equation of the tangent line, we need to:1. Determine \( f'(x) \), the derivative of \( f(x) \).2. Evaluate \( f'(x) \) at \( x = 1 \) to find the slope of the tangent line.3. Use 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. **Step-by-Step Solution:**1. **Differentiate \( f(x) \):** - Use the product rule, where \( u = x^2 \) and \( v = \ln x \). - \( u' = 2x \) and \( v' = \frac{1}{x} \). - Therefore, \( f'(x) = u'v + uv' = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x \ln x + x \).2. **Calculate \( f'(x) \) at \( x = 1 \):** - \( f'(1) = 2(1) \ln 1 + 1 \). - Since \( \ln 1 = 0 \), \( f'(1) = 1 \).3. **Determine \( f(1) \):** - \( f(1) = (1)^2 \ln 1 = 0 \).4. **Equation of the tangent line:** - Using the point \( (1, 0) \) and slope \( m = 1 \), the equation is: \[ y - 0 = 1(x - 1) \] - Simplify to \( y = x - 1 \).Thus, the equation of the tangent line is \( y = x - 1 \).

Name: **Problem Statement:**Find the equation of the tangent line of \( f(x
Uploaded: 2024-03-20T06:47:22.000Z
Channel: Bartleby
Description: **Problem Statement:**Find the equation of the tangent line of \( f(x) = x^2 \ln x \) at \( x = 1 \).**Explanation:**To find the equation of the tangent line, we need to:1. Determine \( f'(x) \), the derivative of \( f(x) \).2. Evaluate \( f'(x) \)