## amy0799 one year ago d/dx m(x) where m(x) is the tangent line to curve x^2+xy-y^2 = 1 at point (2,3)

1. amistre64

the slope of a line is its derivative ...

2. amistre64

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

3. amistre64

right?

4. amy0799

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

5. amistre64

as i see it, yes

6. amy0799

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

7. amistre64

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

8. amistre64

the derivative of a line, is its slope

9. amistre64

|dw:1444585385770:dw|

10. amistre64

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

11. amy0799

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

12. 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

13. amistre64

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

14. amy0799

i got 1.75

15. amistre64

7/4 yes

16. amy0799

thank you!

17. amistre64

good luck