What is the equation of the tangent line of #f(x)=xlnx^2-x^2lnx# at #x=3#?

Question

Isaiah Allen · Accepted Answer

#y=-[1+4ln(3)]x+3[1+3ln(3)]# #f(x)=xln(x^2)-x^2ln(x)# #=(2-x)xln(x)# The equation of the tangent line #x=3# is the same as the first order taylor expansion of #f# about #x=3#. #f(x)~~f(3)+f'(3)(x-3)=y# #f(3)=(2-3)(3)ln(3)# #=-3ln(3)# #f'(x)=frac{d}{dx}((2-x)xln(x))# #=(2-x)xfrac{d}{dx}(ln(x))+(2-x)frac{d}{dx}(x)ln(x)+frac{d}{dx}(2-x)xln(x)# #=(2-x)xfrac{1}{x}+(2-x)(1)ln(x)+(-1)xln(x)# #=2-x+2(1-x)ln(x)# #f'(3)=2-3+2(1-3)ln(3)# #=-1-4ln(3)# #y=-(1+4ln(3))(x-3)-3ln(3)# #=-(1+4ln(3))x+9ln(3)+3#

Ezra Adams · Answer

undefined The equation of the tangent line of f(x)=xlnx^2-x^2lnx at x=3 is y = -3ln3x + 9ln3 - 9.

Ezra Adams · Answer

undefined To find the equation of the tangent line to the function $f(x) = x \ln(x^2) - x^2 \ln(x)$ at $x = 3$, you need to follow these steps:

1. Find the first derivative of $f(x)$ with respect to $x$ to get the slope of the tangent line.
2. Evaluate the derivative at $x = 3$ to find the slope of the tangent line at that point.
3. Use the point-slope form of the equation of a line, substituting the slope and the point $x = 3$ into the equation.

First, find the derivative of $f(x)$:

$$f'(x) = \frac{d}{dx}(x \ln(x^2) - x^2 \ln(x))$$

To find the derivative, you can use the product rule and the chain rule:

$$f'(x) = (1) \cdot \ln(x^2) + x \cdot \frac{1}{x^2} \cdot 2x - x^2 \cdot \frac{1}{x} - 2x \ln(x)$$

Simplify the expression:

$$f'(x) = \ln(x^2) + 2 - \frac{x^2}{x} - 2x \ln(x)$$
$$f'(x) = \ln(x^2) + 2 - x - 2x \ln(x)$$

Next, evaluate the derivative at $x = 3$:

$$f'(3) = \ln(3^2) + 2 - 3 - 2(3) \ln(3)$$
$$f'(3) = \ln(9) + 2 - 3 - 6 \ln(3)$$
$$f'(3) = 2\ln(3) + 2 - 3 - 6 \ln(3)$$
$$f'(3) = -1 - 4 \ln(3)$$

Now, you have the slope of the tangent line at $x = 3$, which is $m = -1 - 4 \ln(3)$.

Using the point-slope form of the equation of a line with the point $(3, f(3))$, substitute the values:

$$y - f(3) = m(x - 3)$$

$$y - (3 \ln(3^2) - 3^2 \ln(3)) = (-1 - 4 \ln(3))(x - 3)$$

$$y - (3 \ln(9) - 9 \ln(3)) = (-1 - 4 \ln(3))(x - 3)$$

$$y - (3 \ln(9) - 9 \ln(3)) = (-1 - 4 \ln(3))x + (3 + 12 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + (3 + 12 \ln(3)) + (3 \ln(9) - 9 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + (3 + 12 \ln(3) + 3 \ln(9) - 9 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + (3 + 12 \ln(3) + 3 \ln(3^2) - 9 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + (3 + 12 \ln(3) + 3 \cdot 2 \ln(3) - 9 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + (3 + 12 \ln(3) + 6 \ln(3) - 9 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + (3 + 9 \ln(3) - 9 \ln(3))$$

$$y = (-1 - 4 \ln(3))x + 3$$

So, the equation of the tangent line to $f(x) = x \ln(x^2) - x^2 \ln(x)$ at $x = 3$ is $y = (-1 - 4 \ln(3))x + 3$.