# What is the equation of the normal line of f(x)= xln(1/x^2) at x = 2?

May 2, 2017

$y = \frac{x}{2 \ln \left(2\right) + 2} - \frac{1}{\ln \left(2\right) + 1} - 4 \ln \left(2\right)$

#### Explanation:

The function $y = x \ln \left(\frac{1}{x} ^ 2\right)$ can be simplified to $y = x \ln \left({x}^{-} 2\right) = - 2 x \ln \left(x\right)$.

A normal line is perpendicular to the tangent line. Thus, we need to find the slope of the tangent line of the function at $x = 2$ (call it $m$). The slope of the normal line of the function at $x = 2$ then is given by $- \frac{1}{m}$. We also know that it crosses $- 2 x \ln \left(x\right)$ at $x = 2 \Rightarrow y = - 2 \cdot 2 \ln \left(2\right) = - 4 \ln \left(2\right)$. Since we know its slope and a point, we can find the equation of the line using the point-slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where $m$ is the slope and $\left({x}_{1} , {y}_{1}\right)$ is a point on the slope.

The slope of the function $y = - 2 x \ln \left(x\right)$ at $x = 2$ can be calculated by using calculus. $\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \ln \left(x\right) - 2$ (using the product rule $\frac{d \left(u v\right)}{\mathrm{dx}} = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$). The slope of the line, $\frac{\mathrm{dy}}{\mathrm{dx}}$, at $x = 2$ is $- 2 \ln \left(2\right) - 2$. Therefore, the slope of the normal line is $- \frac{1}{- 2 \ln \left(2\right) - 2} = \frac{1}{2 \ln \left(2\right) + 2}$.

Therefore, the normal line is given by $y + 4 \ln \left(2\right) = \frac{x - 2}{2 \ln \left(2\right) + 2}$. Simplify this to get $y = \frac{x}{2 \ln \left(2\right) + 2} - \frac{1}{\ln \left(2\right) + 1} - 4 \ln \left(2\right)$.

Verify this using the graph below:
graph{(y-xln(1/x^2))(y-x/(2ln(2)+2)+1/(ln(2)+1)+4ln(2))=0 [-10, 10, -5, 5]}