amy0799
  • amy0799
d/dx m(x) where m(x) is the tangent line to curve x^2+xy-y^2 = 1 at point (2,3)
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
the slope of a line is its derivative ...
amistre64
  • amistre64
d/dx (tangent line to a curve) = y', or slope of the curve
amistre64
  • amistre64
right?

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amy0799
  • amy0799
yes, so I just need to take the derivative of x^2+xy-y^2 = 1?
amistre64
  • amistre64
as i see it, yes
amy0799
  • amy0799
i asked my other friends, and they took the 2nd derivative. so I'm now confused...
amistre64
  • amistre64
\[y_{tan}=f'(a)(x-a)+f(b)\] \[\frac d{dx}y_{tan}=f'(a)\]
amistre64
  • amistre64
the derivative of a line, is its slope
amistre64
  • amistre64
|dw:1444585385770:dw|
amistre64
  • amistre64
regardless of the curve ... the function we are taking the derivative of is a line
amy0799
  • amy0799
\[m'(x)=\frac{ -2x-y }{ x-2y }\] is this correct?
amistre64
  • amistre64
maybe .. now you gonna make me work lol x^2+xy-y^2 = 1 2xx'+x'y+xy' -2yy' = 0 2 3 2 3 4+3+2y' -6y' = 0 (2-6)y' = -7
amistre64
  • amistre64
i tend to have less error when i plug in and then solve for y'
amy0799
  • amy0799
i got 1.75
amistre64
  • amistre64
7/4 yes
amy0799
  • amy0799
thank you!
amistre64
  • amistre64
good luck

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