Find the equation of the tangent to the curve $\displaystyle y = \frac {x^2}{x^2 + 1}$ at $\displaystyle x = 1 $

$\displaystyle y' = \frac {2x}{(x^2 + 1)^2} $

$\displaystyle f'(1) = \frac {1}{2} $

$\displaystyle f(1) = \frac {1}{2} $

$\displaystyle \frac {1}{2} = \frac {1}{2} x + b$

$\displaystyle 0 = b$

Therefore, $\displaystyle y = \frac {1}{2} x$

Textbook Answer:$\displaystyle x - 2y = 0 $.... what did I do wrong?