Question

implicit differentiation

Answer 1

\[x^{2}-xy+y^{2} = 27\]

\[(d/dx)x^{2}-xy+y^{2} = (d/dx)27\]


\[2x-xy'+y+2y' = 0\]

\[-y'(x-2y)+(2x+y) = 0\]

\[\frac{ (2x+y) }{ (x-2y) } = y'\]

Answer 2

my final answer is wrong btw

Answer 3

\[\dfrac{d}{dx} y^2 = 2 y y'\]

Answer 4

\[2x-xy'+y+2yy' = 0\]

Answer 5

y

Answer 6

unless.. we take -x as u and v as y

u = -x du = -1
v = y dv = y'

-xy'-y

Answer 7

I'm also really bad when it comes to negatives terms..

Answer 8

\[\checkmark\]

Answer 9

okay cool.

Answer 10

lol accidently deleted my response

Answer 11

so how would I group the terms properly? @IrishBoy123

Answer 12

i'd first take my time....

i think we had \((x^{2}-xy+y^{2} = 27)' \implies 2x - y - x y' + 2 y y' = 0\)

so \( y' ( 2 y- x ) =y - 2x\)

etc

Answer 13

so \( \dfrac{dy}{dx} = \dfrac{y - 2x}{2y - x} \)


from there, the idea that \(p(x) = \frac{y}{x}\) makes sense in terms of trying to grab a solution

Answer 14

did that make sense????

cos you can really branch out

\[f(x,y) = x^{2}-xy+y^{2} = 27\]

\[f_x = 2x - y \]
\[f_y = -x+2y \]

\[ \dfrac{dy}{dx} = -\dfrac{f_x}{f_y} = \dfrac{2x - y}{-x+2y} \]

Answer 15

thanks