SOLUTION: implicit differeniation to find y' and evaluate y' at the indicated point . 1) xIn y + 2y= 2x^3 : (1,1) 2) x^3 - tx^2 - 4= 0 :(-3,-2)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: implicit differeniation to find y' and evaluate y' at the indicated point . 1) xIn y + 2y= 2x^3 : (1,1) 2) x^3 - tx^2 - 4= 0 :(-3,-2)      Log On


   

Question 384079: implicit differeniation to find y' and evaluate y' at the indicated point .
1) xIn y + 2y= 2x^3 : (1,1)

2) x^3 - tx^2 - 4= 0 :(-3,-2)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%2Aln%28y%29+%2B+2y=+2x%5E3
x%2A%28dy%2Fy%29%2Bln%28y%29%2Adx%2B2dy=6x%5E2%2Adx
%28x%2Fy%2B2%29dy=%286x%5E2-ln%28y%29%29dx
dy%2Fdx=%286x%5E2-ln%28y%29%29%2F%28x%2Fy%2B2%29
When x=y=1
dy%2Fdx=%286-ln%281%29%29%2F%281%2F1%2B2%29
dy%2Fdx=%286%29%2F%283%29
dy%2Fdx=2
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+x%5E3+-tx%5E2+-+4=+0+
3x%5E2%2Adx-%28t%282x%2Adx%29%2Bx%5E2%2Adt%29=0
3x%5E2%2Adx-2xt%2Adx-x%5E2%2Adt=0
%283x%5E2-2xt%29%2Adx=x%5E2%2Adt
dx%2Fdt=x%5E2%2F%283x%5E2-2xt%29
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If t=-3 and x=-2, then,
dx%2Fdt=4%2F%2812-12%29
dx%2Fdt is undefined.
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If t=-2 and x=-3, then,
dx%2Fdt=9%2F%2827-12%29
dx%2Fdt=9%2F15
dx%2Fdt=3%2F5