# Thread: Derivative of a Function at a specific point

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

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

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

Here's what I did:

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

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

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

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

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

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

Here's what I did:

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

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

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

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

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