implicit differentiation: y^2 - 2xy = 11

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Laila

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Jul 10, 2006
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ok, the question is.....
find the slope of the tangent to the curve and the given point
y^2 - 2xy = 11 (5, -1)


this is what i did so far....
y^2 - 2xy = 11
2y (dy/dx) - 2y - (dy/dx)2x =0
dy/dx (2y -2x) - 2y =0
dy/dx = 2y/2y -2x
= 1/-2x
the slope at point (5, -1) is

1/-10

this is the answer i got but its wrong, its supposed to be 1/6

I just dont see what im doing wrong. there are a lot of questions asking
the same thing and i follow the same format to answer them but i keep
getting them wrong. i just dont see what im doing wrong. can anyone help me
please?
 
Re: implicit differentiation

Laila said:
ok, the question is.....
find the slope of the tangent to the curve and the given point
y^2 - 2xy = 11 (5, -1)


this is what i did so far....
y^2 - 2xy = 11
2y (dy/dx) - 2y - (dy/dx)2x =0
dy/dx (2y -2x) - 2y =0
dy/dx = 2y/(2y-2x) = y/(y-x), now sub in your values ... you'll get 1/6
= 1/-2x
the slope at point (5, -1) is

1/-10

this is the answer i got but its wrong, its supposed to be 1/6

I just dont see what im doing wrong. there are a lot of questions asking
the same thing and i follow the same format to answer them but i keep
getting them wrong. i just dont see what im doing wrong. can anyone help me
please?
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