$\displaystyle 2y^2-2x^2+0=0$
evaluate y' at given point (-2,-2)
What is this called?
How would I do it?
The relation $\displaystyle y^{2} - x^{2}=0$ defines in implicit form a function $\displaystyle y= f(x)$ and its derivative in a point $\displaystyle (y_{0},x_{0})$ can be obtained from a well known explicit formula...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
If You have a function $\displaystyle y(x)$ implicity defined as...
$\displaystyle f(x,y)=0$ (1)
... its derivative in a point $\displaystyle (x_{0},y_{0})$ is...
$\displaystyle y^{'} (x_{0}) = - \frac{f^{'}_{x} (x_{0},y_{0})}{f^{'}_{y} (x_{0},y_{0})}$ (2)
In Your case is $\displaystyle f(x,y)= y^{2} - x^{2}$, $\displaystyle x_{0}=-2$, $\displaystyle y_{0}=-2$, so that is $\displaystyle y^{'} (-1) = 1$. A more simple solution for this particular case is to write (1) as $\displaystyle y^{2}=x^{2}$ so that $\displaystyle y(x)= \pm x$. The 'right' $\displaystyle y(x)$ passes from the point $\displaystyle (-2,-2)$ so that is $\displaystyle y(x)=x$ and $\displaystyle y^{'} (x)=1$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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