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) About Me  (Show Source):
You can put this solution on YOUR website!
Use implicit differentiation,
3x%5E2dx%2B4%28xdy%2Bydx%29-2ydy=0
3x%5E2dx%2B4xdy%2B4ydx-2ydy=0
%284x-2y%29dy=-%283x%5E2%2B4y%29dx
dy%2Fdx=-%28%283x%5E2%2B4y%29%2F%284x-2y%29%29
So when x=2 and y=1,
dy%2Fdx=-%28%283%282%29%5E2%2B4%281%29%29%2F%284%282%29-2%281%29%29%29
dy%2Fdx=-%28%2812%2B4%29%2F%288-2%29%29
dy%2Fdx=-%2816%2F6%29
dy%2Fdx=-%288%2F3%29
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Answer by Edwin McCravy(20056) About Me  (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