Question 1166986
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3xy^2\ +\ 2y\ -\ 3\ =\ 2]


Implicit differentiation, presuming *[tex \Large y] is a function of *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\cdot\frac{d(y^2)}{dy}\cdot\frac{dy}{dx}\ +\ \frac{d(3x)}{dx}\cdot y^2\ +\ \frac{d(2y)}{dy}\cdot\frac{dy}{dx}\ -\ \frac{d(3)}{dx}\ =\ \frac{d(2)}{dx}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6xy\frac{dy}{dx}\ +\ 3y^2\ +\ 2\frac{dy}{dx}\ -\ 0\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}(6xy\ +\ 2)\ =\ -3y^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'\ =\ \frac{dy}{dx}\ =\ -\frac{3y^2}{6xy\ +\ 2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'\|_{(1,1)}\ =\ \frac{3}{8}]


For the *[tex \Large y'\text{'}] part you need to implicitly differentiate *[tex \Large y'] with respect to *[tex \Large x]. You will need the Quotient Rule, the Product Rule, and the Chain Rule.  Good luck and have fun.

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


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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