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Find y' and the slope of the tangent line to the graph of the equation: 6sq...
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`6sqrt(y^3+1)-2x^(3/2) - 2 = 0`
Using implicit differentiation,
`6 xx1/2 (y^3+1)^(-1/2) xx 3y^2 xxdy - 2 xx 3/2 xx x^(1/2)dx - 0 = 0`
`(9y^2)/sqrt(y^3+1)xxdy- 3sqrt(x)dx = 0`
`(dy)/(dx) = (3sqrt(x))/((9y^2)/sqrt(y^3+1))`
`(dy)/(dx) = (sqrt(x)sqrt(y^3+1))/(3y^2)`
`(dy)/(dx) = (sqrt(4)sqrt(3^3+1))/(3xx3^2)`
`(dy)/(dx) = (2sqrt(10))/(27)`
Therefore the slope of the tangent line to the graph at (4,3) is `(2sqrt(10))/(27)`
Posted by thilina-g on June 18, 2012 at 10:19 AM (Answer #1)
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