Sunday

August 31, 2014

August 31, 2014

Posted by **Ali** on Sunday, November 18, 2012 at 2:20pm.

Help please? Thanks in advance

- Calculus [finding slope of tangent line] -
**Steve**, Sunday, November 18, 2012 at 2:27pmDon't know why you bother with the zero:

−2xy+2y^3=10

(yes, (-4,1) is on that graph.

just use the product rule, remembering that x' = 1:

-2yx' - 2xy' + 6yy' = 0

-2y + (6y-2x)y' = 0

y' = 2y/(6y-2x)

at (-4,1),

y' = 2(1)/(6(1)-2(-4)) = 2/(6+8) = 1/7

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