# Math Help - Derivative of a Function at a specific point

1. ## Derivative of a Function at a specific point

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

1. (A) 2
2. (B) $\frac{1}{2}$
3. (C) 0
4. (D) $\frac{-1}{2}$
5. (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}
$

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

1. (A) 2
2. (B) $\frac{1}{2}$
3. (C) 0
4. (D) $\frac{-1}{2}$
5. (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'

...

3. Ohh... Okay, so then,

I get y' = -3/2 ?

But that's not right...

4. Originally Posted by StarlitxSunshine
Ohh... Okay, so then,

I get y' = -3/2 ?

But that's not right...
your mistake (and mine) ...

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

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

Thank you :33

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