If $\displaystyle 4x^2-4xy+y^2=1$ then $\displaystyle \frac{dy}{dx}$ at $\displaystyle (1,1)$ =

- (A) 2
- (B) $\displaystyle \frac{1}{2}$
- (C) 0
- (D) $\displaystyle \frac{-1}{2}$
- (E)$\displaystyle -2$

Here's what I did:

$\displaystyle

\frac{dy}{dx} [4x^2-4xy+y^2] = \frac{dy}{dx} [1]

$

$\displaystyle

8x - 4((x)(y')+(y)(1)) + 2yy' = 1

$

$\displaystyle

8 -4(y' + 1) + 2y' = 1

$

$\displaystyle

y' = \frac {-11}{-2}

$