Question 1193305
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\[x^2\]\ =\ 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\[y^2\]\ =\ 2y\cdot\frac{dy}{dx}\ \ \ \ ]  (Chain Rule)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\[9\]\ =\ 0,\ \ \ \ ] so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\[x^2\,+\,y^2\,=\,9\]\ \equiv\ 2x\ + \ 2y\cdot\frac{dy}{dx}\ =\ 0]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2y\frac{dy}{dx}\ =\ -2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ -\frac{x}{y}]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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