which now looking at it should be 2
general
Phi Kappa Chi ΦΚΧ <a:yeah:589703000977571860><a:ppHop:536353159615086612>
Wednesday, October 25, 2017
yes thats the real question
like i said i forgot the = 9
regardless
all i need
alright
is the double derivative of x^2-xy+y^2=9
I remebered how to do this
the single derivative is (2x-y)/(x-2y)
dy/dx = (y-2x)/(1+2y)
i can visually see that it is 2
on the graph
or x -2y
you were right all along, y is dependant on x leg
because they do not actually need to solve it
its ab calc
and yeah
i think
I had to do this too, it is not actually bad
and I never took any advanced shit
its not bad
im just confused as shit on taking the derivative of dy/dx = (y-2x)/(1+2y)
literally product rule, and chain rule, then rearange for dy/dx
do it the same way
product rule?
quotient rule rather
oh yeah
wait chain rule?
where can i use chain rule
and take derivatives of y as if they were x, then multiply it by dy/dx
i know the chain rule
then sub in dy/dx that you solved for before
yeah i tried that
that is what you are doing when you take a derivative of y and make it dy/dx
and ended up with a mess of shit
should be a mess of shit
give me a sex
sec
fuck
