Question

if xy + y = 3, then dy/dx =

Answer 1

Implicit Differentiation

Answer 2

\[xy+y=3\]\[y(x+1)=3\]\[y=\frac3{x+1}\]

Answer 3

the answers are -y/1+x, y/1+x, 3/y, 3/1+y, and 3/1-y

Answer 4

\[y=\frac3{x+1}\]
\[\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\frac3{x+1}\]\[\qquad=\frac{\mathrm d}{\mathrm dx}\left(3\times({x+1})^{-1}\right)\]
\[\qquad=\frac{\mathrm d}{\mathrm dx}\left(3\right)\times(x+1)^{-1}+3\times\frac{\mathrm d}{\mathrm dx}\left((x+1)^{-1}\right)\]

Answer 5

what is the answer then?

Answer 6

xy + y = 3

y + x dy/dx + dy/dx = 0

(1 +x) dy/dx = -y

dy/dx = - y / (1 + x)

Answer 7

a bit easier if u Use Implicit Differentiation

Answer 8

thanks so much!!!!

Answer 9

@ammubhave u got it

Answer 10

Welcome..

Answer 11

\[\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\left(3\right)\times(x+1)^{-1}-3\times\frac{\mathrm d}{\mathrm dx}\left((x+1)^{-1}\right)\]\[=0-3\times(x+1)^{-2}\]\[=\frac{-3}{(x+1)^2}\]\[=\frac{-1}{x+1}\times\frac3{x+1}\]\[=\frac{-1}{x+1}\times y\]\[=\frac {-y}{x+1}\]