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

Question

amistre64 · Answer

the slope of a line is its derivative ...

amistre64 · Answer

d/dx (tangent line to a curve) = y', or slope of the curve

amistre64 · Answer

right?

amy0799 · Answer

yes, so I just need to take the derivative of x^2+xy-y^2 = 1?

amistre64 · Answer

as i see it, yes

amy0799 · Answer

i asked my other friends, and they took the 2nd derivative.  so I'm now confused...

amistre64 · Answer

$$y_{tan}=f'(a)(x-a)+f(b)$$
$$\frac d{dx}y_{tan}=f'(a)$$

amistre64 · Answer

the derivative of a line, is its slope

amistre64 · Answer

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amistre64 · Answer

regardless of the curve ... the function we are taking the derivative of is a line

amy0799 · Answer

$$m'(x)=\frac{ -2x-y }{ x-2y }$$
is this correct?

amistre64 · Answer

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 · Answer

i tend to have less error when i plug in and then solve for y'

amy0799 · Answer

i got 1.75

amistre64 · Answer

7/4  yes

amy0799 · Answer

thank you!

amistre64 · Answer

good luck