The function y=xln(1/x^2) can be simplified to y=xln(x^-2)=-2xln(x).
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 -1/m. We also know that it crosses -2xln(x) at x=2rArry=-2*2ln(2)=-4ln(2). 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(x-x_1), where m is the slope and (x_1,y_1) is a point on the slope.
The slope of the function y=-2xln(x) at x=2 can be calculated by using calculus. dy/dx=-2ln(x)-2 (using the product rule (d(uv))/dx=v(du)/dx+u(dv)/dx). The slope of the line, dy/dx, at x=2 is -2ln(2)-2. Therefore, the slope of the normal line is -1/(-2ln(2)-2)=1/(2ln(2)+2).
Therefore, the normal line is given by y+4ln(2)=(x-2)/(2ln(2)+2). Simplify this to get y=x/(2ln(2)+2)-1/(ln(2)+1)-4ln(2).
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]}
