$\displaystyle x^2-xy-4y^2=0$
Find: $\displaystyle (dy)/(dx)$
Thanks in advance!
I got this answer:
$\displaystyle
(dy)/(dx)=(y+10x)/(x+8y)
$
and i just want to make sure I did it right...
$\displaystyle x^2-xy-4y^2=0$
Find: $\displaystyle (dy)/(dx)$
Thanks in advance!
I got this answer:
$\displaystyle
(dy)/(dx)=(y+10x)/(x+8y)
$
and i just want to make sure I did it right...
Edit: Oh you just had to get banned. =p
$\displaystyle x^2 - xy - 4y^2 = 0$
$\displaystyle (2x) - (y + xy') - (8yy') = 0$
$\displaystyle 2x - y - xy' - 8yy' = 0$
$\displaystyle (xy' + 8yy') = (2x - y)$
$\displaystyle y'(x + 8y) = (2x - y)$
$\displaystyle y' = \frac{2x - y}{x + 8y}$
Easier way to do it when you've got it in that form: -(d/dx)/(d/dy)
d/dx = 2x - y
d/dy = -x - 8y
$\displaystyle \frac{-(d/dx)}{(d/dy)} = \frac{-(2x - y)}{-x - 8y} = \frac{2x - y}{x + 8y}$