Question

d/dx m(x) where m(x) is the tangent line to curve x^2+xy-y^2=1 at point (2,3)

Answer 1

my work:
\(
2x+y+xy'-2yy'=0\\
2x+y=2yy'-xy'\\
2x+y=y'(2y-x)\\
\dfrac{dy}{dx}=\dfrac{2x+y}{2y-x}
\)

Answer 2

do i just plug the point in now? or do I use y-y1=m(x-x1) somehow

Answer 3

Since m'(x)=the slope m at (2,3),
y'(2) should equal m'(x).

Answer 4

* y'(2,3)

Answer 5

I'm not sure I understand what the question is asking? is 7 the answer?

Answer 6

\(\dfrac{d}{dx}(m(x))\) means : after you got m(x), take derivative of it again.

Answer 7

ohhh

Answer 8

and then plug (2,3) in

Answer 9

it's been like 2 years since I took a calc class, was helping someone with this question.

thank you both

Answer 10

thank you https://www.desmos.com/calculator/ykoruebu4u

Answer 11

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)