anonymous
  • anonymous
d/dx m(x) where m(x) is the tangent line to curve x^2+xy-y^2=1 at point (2,3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
my work: \( 2x+y+xy'-2yy'=0\\ 2x+y=2yy'-xy'\\ 2x+y=y'(2y-x)\\ \dfrac{dy}{dx}=\dfrac{2x+y}{2y-x} \)
anonymous
  • anonymous
do i just plug the point in now? or do I use y-y1=m(x-x1) somehow
mathmate
  • mathmate
Since m'(x)=the slope m at (2,3), y'(2) should equal m'(x).

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mathmate
  • mathmate
* y'(2,3)
anonymous
  • anonymous
I'm not sure I understand what the question is asking? is 7 the answer?
Loser66
  • Loser66
\(\dfrac{d}{dx}(m(x))\) means : after you got m(x), take derivative of it again.
anonymous
  • anonymous
ohhh
Loser66
  • Loser66
and then plug (2,3) in
anonymous
  • anonymous
it's been like 2 years since I took a calc class, was helping someone with this question. thank you both
anonymous
  • anonymous
thank you https://www.desmos.com/calculator/ykoruebu4u
mathmate
  • mathmate
|dw:1444436902647:dw| The question is asking for the slope of the tangent line m(x), which should be the same as the slope of the curve at (2,3), by definition of the tangent. that is why m'(x)=slope of tangent line = y'(2,3)

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