SOLUTION: Find the gradient of the curve x^3+4xy-y^2=15 at the point (2,1)

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Question 990509: Find the gradient of the curve x^3+4xy-y^2=15 at the point (2,1)
Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Use implicit differentiation,




So when and ,




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Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Find the gradient of the curve x^3+4xy-y^2=15 at the point (2,1)

x³ + 4xy - y² = 15

3x² + 4(xy'+ y) - 2yy' = 0

3x² + 4xy' + 4y - 2yy' = 0

3x² + 4y = 2yy' - 4xy'

3x² + 4y = 2y'(y-2x)

3x² + 4y
———————— = y'
2(y-2x) 

Substitute (2,1)

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

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

  12 + 4
—————————— = y'(2)
   2(-3)

 16
———— = y'(2)
 -6

   8
- ——— = y'(2)
   3

Edwin
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