1. ## Intro to Derivatives

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?

2. Originally Posted by Macleef
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?
Um...nothing

y = (1/2)x

=> 2y = x

=> 0 = x - 2y

the textbook just has the equation in standard form, that's all