Question

Find dy/dx by implicit differentiation
(x+y)/(x-y)=x^2+y^2

Answer 1

Sorry not to be a wingspan to anyone that is going to answer but don't use wolframalpha or mathway since they are wrong.

Answer 2

\(\dfrac{x+y}{x-y}=x^2+y^2\)

\( \dfrac{(x - y)(1 + y') - (x + y)(1 - y')}{(x - y)^2} = 2x + 2yy' \)

Answer 3

Now you need to muliply out the left side and isolate y'.

Answer 4

x+y = (x^2 + y^2)(x - y)
x + y = x^3 - yx^2 + xy^2 - y^3
\[\frac{ dy }{ dx }=3x ^{2}-\frac{ dy }{ dx }x ^{2}-2yx+y ^{2}+2xy \frac{ dy }{ dx }-3y ^{2}\frac{ dy }{ dx }\]
\[\frac{ dy }{ dx }+\frac{ dy }{ dx }x ^{2}-2xy \frac{ dy }{ dx }+3y ^{2}\frac{ dy }{ dx }=3x ^{2}-2yx+y ^{2}\]
\[\frac{ dy }{ dx }(1+x ^{2}-2xy +3y ^{2})=3x ^{2}-2yx+y ^{2}\]

Answer 5

and for some reason the functions are working so the next step is to devide both sides by 1 + x^2 - 2xy + 3y^2

Answer 6

Above, from the 2nd line to the 3rd line, you took the derivative of x + y on the left side.
Shouldn't it be 1 + y'?

Answer 7

yes it should, sry my bad, so wat i meant to do was put the 1 on the right side after wards and then divide both side by 1+ x^2 - 2xy + 3y^2

Answer 8

ill fix it now quickly lol

Answer 9

x+y = (x^2 + y^2)(x - y)
x + y = x^3 - yx^2 + xy^2 - y^3
\[1+\frac{ dy }{ dx }=3x ^{2}-\frac{ dy }{ dx }x ^{2}-2yx+y ^{2}+2xy \frac{ dy }{ dx }-3y ^{2}\frac{ dy }{ dx }\]
\[\frac{ dy }{ dx }+\frac{ dy }{ dx }x ^{2}-2xy \frac{ dy }{ dx }+3y ^{2}\frac{ dy }{ dx }=3x ^{2}-2yx+y ^{2}-1\]
\[\frac{ dy }{ dx }(1+x ^{2}-2xy +3y ^{2})=3x ^{2}-2yx+y ^{2}-1\]

Answer 10

\(\dfrac{(x - y)(1 + y') - (x + y)(1 - y')}{(x - y)^2} = 2x + 2yy'\)

\(\dfrac{x + xy' - y - yy' - x + xy' - y +yy'}{(x - y)^2} = 2x + 2yy'\)

\(xy' - y + xy' - y = (x - y)^2( 2x + 2yy')\)

\(2xy' - 2y = 2x(x - y)^2 + 2yy'(x - y)^2\)

\(2xy' - 2yy'(x - y)^2 = 2x(x - y)^2 + 2y\)

Answer 11

im just happy ik how to do this question ^^

Answer 12

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

\( y' = \dfrac{ 2x(x - y)^2 + 2y }{ 2x - 2y(x - y)^2 } \)

\( y' = \dfrac{ x(x - y)^2 + y }{ x - y(x - y)^2 } \)