Answer
$$y = x + 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{2}x\ln {x^2},{\text{ }}\left( { - 1,0} \right) \cr
& {\text{Differentiate }}f\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}x\ln {x^2}} \right] \cr
& f'\left( x \right) = \frac{1}{2}x\left( {\frac{{2x}}{{{x^2}}}} \right) + \ln {x^2}\left( {\frac{1}{2}} \right) \cr
& f'\left( x \right) = 1 + \frac{1}{2}\ln {x^2} \cr
& {\text{Calculate the slope at the given point }}\left( { - 1,0} \right) \cr
& m = f'\left( { - 1} \right) = 1 + \frac{1}{2}\ln {\left( { - 1} \right)^2} \cr
& m = 1 \cr
& {\text{Find the equation of the tangent line at }}\left( { - 1,0} \right) \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 0 = \left( {x + 1} \right) \cr
& y = x + 1 \cr
& \cr
& {\text{Graph}} \cr} $$
