Question

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

Answer 1

I just started learning this too :) lurking to see what I can learn.. there are some great helpers on tonight. hold tight

Answer 2

take them on side first

Answer 3

do you know how to use the quotient rule?

Answer 4

\[f(x,y)=x^2-\frac{x+y}{x-y}=0\\y'=-\frac{ f'_x(x,y) }{ f'_y(x,y) }\]

Answer 5

\[2x-\frac{ (x+y) }{ (x-y )}=0\]\[2x-\frac{ (x-y)(1+\frac{ dy }{ dx })-(x+y)(1-\frac{ dy }{ dx }) }{ (x-y)^{2} }=0\]\[2x-[(x+x \frac{ dy }{ dx }-y-y \frac{ dy }{ dx })- (x-x \frac{ dy }{ dx }+y-y \frac{ dy }{ dx }]/(x-y)^{2}=0\]\[2x-\frac{ x+x \frac{ dy }{ dx }-y-y \frac{ dy }{ dx } -x+x \frac{ dy }{ dx }-y+y \frac{ dy }{ dx }}{ (x-y )^{2}}=0\]\[2x-\frac{ 2x \frac{ dy }{ dx }-2y }{ (x-y)^{2}}=0\]\[2x=\frac{( 2x \frac{ dy }{ dx }-2y) }{ (x-y)^{2} }\]\[2x(x-y)^{^{2}}=2x \frac{ dy }{ dx }-2y\]\[2x(x-y)^{2}+2y=2x \frac{ dy }{ dx }\]\[\frac{ 2x(x-y)^{2}+2y }{ 2x }=\frac{ dy }{ dx }\]\[(x-y)^{2}+\frac{ y }{ x }=\frac{ dy }{ dx }\]