# Thread: What is this called?

1. ## What is this called?

$2y^2-2x^2+0=0$

evaluate y' at given point (-2,-2)

What is this called?

How would I do it?

2. Originally Posted by Zanderist
$2y^2-2x^2+0=0$

evaluate y' at given point (-2,-2)

What is this called?

How would I do it?
Go to your notes or textbook and look for the heading Implicit Differentiation.

3. The relation $y^{2} - x^{2}=0$ defines in implicit form a function $y= f(x)$ and its derivative in a point $(y_{0},x_{0})$ can be obtained from a well known explicit formula...

Kind regards

$\chi$ $\sigma$

4. So basically all I have to do is isolate y, then find the derivative?

5. If You have a function $y(x)$ implicity defined as...

$f(x,y)=0$ (1)

... its derivative in a point $(x_{0},y_{0})$ is...

$y^{'} (x_{0}) = - \frac{f^{'}_{x} (x_{0},y_{0})}{f^{'}_{y} (x_{0},y_{0})}$ (2)

In Your case is $f(x,y)= y^{2} - x^{2}$, $x_{0}=-2$, $y_{0}=-2$, so that is $y^{'} (-1) = 1$. A more simple solution for this particular case is to write (1) as $y^{2}=x^{2}$ so that $y(x)= \pm x$. The 'right' $y(x)$ passes from the point $(-2,-2)$ so that is $y(x)=x$ and $y^{'} (x)=1$...

Kind regards

$\chi$ $\sigma$

6. Originally Posted by Zanderist
$2y^2-2x^2+0=0$

evaluate y' at given point (-2,-2)

What is this called?

How would I do it?
Implicit diferentiation gives you $4y \frac{dy}{dx} - 4x = 0$. Substitute (-2, -2) and solve for dy/dx.

7. I've manged to get the answer 1.

But that was only a lucky guess steming from the research I did on this website.

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

8. Originally Posted by Zanderist
I've manged to get the answer 1. [snip]
Correct.