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Could you please help me with this calculus problem?

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Using implicit differenctiation find the slope of the curve xy^3+xy=8 at the point (4,1)

asked Nov 18, 2014 in CALCULUS by anonymous

1 Answer

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The curve equation xy3 + xy = 8

Differentiate implicitly with respect to x.

3xy2y' + y3 + xy'+ y = 0

3xy2y'+ xy' = - y3 - y

y'(3xy2+ x) = - y3 - y

y' = (- y3 - y)/(3xy2+ x)

Substitute the values (x , y ) = (4, 1) in above equation.

y' = (- 13 - 1)/[3(4)(1)2 + 4]

y' = (- 1 - 1)/[12 + 4]

y' = -2/16

y' = -1/8

(-1/8) is the slope of tangent line to the curve at (4, 1).

answered Nov 18, 2014 by david Expert

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