1. ## implicit differentiation

I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1

2. Originally Posted by kwivo
I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1
Implicit differentiation is really just the chain rule in disguise.

If you have a function f(y) and take y = y(x) then you've got f(y(x)). Take the x derivative and you get f'(y(x)) y'(x) by the chain rule. This is exactly the same thing as saying
$\frac{d}{dx}f(y) = f^{\prime}(y) \cdot \frac{dy}{dx}$

$4x^2 + 4xy - y^2 = 1$

Take the derivative:
$4 \cdot 2x + \left ( 4y + 4x \frac{dy}{dx} \right ) - 2y \cdot \frac{dy}{dx} = 0$

Now solve for $\frac{dy}{dx}$

-Dan

3. Originally Posted by kwivo
I don't really understand what to do for implicit differentiation. For example, how do I go about doing this:

Find dy/dx by implicit differentiation. 4x2 + 4xy - y2 = 1
for implicit differentiation, you just find the derivative of each variable, taking account for what you are differentiating and with respect to what.

if you differentiate an x-term you attach dx/dx to it (since you took the derivative of the variable x with respect to x), we usually don't write it though, because derivative notations can function as fractions and so dx/dx cancels to become 1

similarly, when differentiating a y-term, we attach dy/dx to it, since we took the derivative of y with respect to x. we often write y' when it is understood what we are differentiating with respect to

here goes:

$4x^2 + 4xy - y^2 = 1$

differentiating implicitly, we get:

$8x + 4y + 4x~y' - 2y~y' = 0$

now solve for $y'$, which is $\frac {dy}{dx}$

EDIT: Beaten by "the man." And he gave a better explanation too!

4. One more question: I don't understand how you get the dy/dx after the 2y. Where did that come from? Isn't 2y already the derivative of y^2.

5. did you read the posts? what did you not get?
Originally Posted by topsquark
If you have a function f(y) and take y = y(x) then you've got f(y(x)). Take the x derivative and you get f'(y(x)) y'(x) by the chain rule. This is exactly the same thing as saying
$\frac{d}{dx}f(y) = f^{\prime}(y) \cdot {\color{red}\frac{dy}{dx}}$
Originally Posted by Jhevon
for implicit differentiation, you just find the derivative of each variable, taking account for what you are differentiating and with respect to what.

when differentiating a y-term, we attach dy/dx to it, since we took the derivative of y with respect to x.

6. Sorry, I overlooked that part.