Equation of tangent line to x^2 + 2xy - y^2 + x = 2 at point (1, 2)

[frank789]

Hi all

so given x^2 + 2xy - y^2 + x = 2 and the point (1,2)

asked for the tangent line. I started doing implicit differentiation but once i got the derivative,

(2(x+y) + 1) / 2(y-x) = y'

can i evaluate y'(1,2)

(2(1+2)+1) / 2(2-1) = (slope?)

i think im right with the equation of

y = 7x/2 - 3/2

verification?
furthermore, if you wouldnt mind helping me shore up my understanding

when i do d/dx of y or y^2 in implicit diff, i know i get y' and 2y(y') respectively. d/dx y is just the definition of y' and the 2y(y') comes from chain rule? (forgive my switching of notation i guess it just makes sense in my head)

lastly, when i do d/dx of y how come i can factor this to (1/y)(y')? i dont understand why that works. mostly likely im missing a key concept in the definition of the method im using.

just a side note its kind of funny when you think about year to year the increase in complexity of what you personally consider "1 line worth of math" i.e factoring was a few lines a couple years ago and if you cant do it in your head at this point your in trouble. can only imagine when (if) i make it the level you guys are at

 

[tkhunny]

You have it right. Why do you doubt?

y = f(x)

\(\displaystyle \dfrac{d}{dx}y = f'(x) = y'\)

\(\displaystyle \dfrac{d}{dx}y^{2} = 2\cdot f(x) \cdot f'(x) = 2yy'\)

The derivatives are WRT x. y will be implicit.

 

[frank789]

ok thanks