Asked by David
If 3x+xy+4y=8, what is the value of d^2y/dx^2 at the point (1,1)? My work:
3+(x)(dy/dx)+(y)+4(dy/dx)=0
dy/dx(x+4)=-3-y
dy/dx=-3-y/x+4
2nd Dervative: (x+4)(0-dy/dx)-(-3-y)/(x+4)^2
(x+4)(-(-y-3/x+4))-(-3-y)/(x+4)^2
(x+4)(y+3)/(-x+4)+(y+3)/(x+4)^2
*Mult. everything by (-x+4)
Simplify:
(x+4)(y+3)+(y+3)(-x+4)/(x+4)^2(-x+4)
*The (-x+4) cancels
Left over: (x+4)(y+3)+(y+3)/(x+4)^2
*Plug in (1,1)
Hi I know how to do this problem, but can you see if i did it right?
3+(x)(dy/dx)+(y)+4(dy/dx)=0
dy/dx(x+4)=-3-y
dy/dx=-3-y/x+4
2nd Dervative: (x+4)(0-dy/dx)-(-3-y)/(x+4)^2
(x+4)(-(-y-3/x+4))-(-3-y)/(x+4)^2
(x+4)(y+3)/(-x+4)+(y+3)/(x+4)^2
*Mult. everything by (-x+4)
Simplify:
(x+4)(y+3)+(y+3)(-x+4)/(x+4)^2(-x+4)
*The (-x+4) cancels
Left over: (x+4)(y+3)+(y+3)/(x+4)^2
*Plug in (1,1)
Hi I know how to do this problem, but can you see if i did it right?
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