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how do you find the slope of the tangent ?
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take the derivative, then plug in 1
thats it?!
thanks!
well take the derivative then plug in (1,1) your derivative will contain y's and x's
is it -5/4 then?
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-1.25 in decimal it was correct ty!
\(x^3+2xy+y^2=4 at (1,1) \(3x^2+2(y+xy')+2yy'=0\\3x^2+2y+2xy'+2yy'=0\\y'=\frac{-3x^2-2y}{2x+2y}\) \) yes
:) thanks for writing that out,
np
wow my latex skills failed me, but you got it without it anyway:)
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