# Intro to Derivatives

• May 7th 2008, 06:46 PM
Macleef
Intro to Derivatives
Find the equation of the tangent to the curve $y = \frac {x^2}{x^2 + 1}$ at $x = 1$

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

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

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

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

$0 = b$

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

Textbook Answer: $x - 2y = 0$ .... what did I do wrong?
• May 7th 2008, 06:54 PM
Jhevon
Quote:

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

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

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

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

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

$0 = b$

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

Textbook Answer: $x - 2y = 0$ .... 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