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

the slope of a line is its derivative ...

Answer 2

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

Answer 3

right?

Answer 4

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

Answer 5

as i see it, yes

Answer 6

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

Answer 7

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

Answer 8

the derivative of a line, is its slope

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

i got 1.75

Answer 14

7/4 yes

Answer 15

thank you!

Answer 16

good luck