Derivative of a Function at a specific point

**StarlitxSunshine** · November 8th 2009, 11:49 AM

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

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

Here's what I did:

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

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

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

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

**skeeter** · November 8th 2009, 11:57 AM

Originally Posted by StarlitxSunshine

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

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

Here's what I did:

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

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

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

... $\textcolor{red}{8 - 4y' - 4 + 2y' = 1}$

recalculate y'

...

**StarlitxSunshine** · November 8th 2009, 12:08 PM

Ohh... Okay, so then,

I get y' = -3/2 ?

But that's not right...

**skeeter** · November 8th 2009, 12:20 PM

Originally Posted by StarlitxSunshine

Ohh... Okay, so then,

I get y' = -3/2 ?

But that's not right...

your mistake (and mine) ...

$8 - 4y' - 4 + 2y' = \textcolor{red}{0}$

**StarlitxSunshine** · November 8th 2009, 12:30 PM

Oh! >_< Wow, I can't believe I did that. x3

Thank you :33

Now, I get -4/-2, which is 2, which is (A). =)

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