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How do you find #(dy)/(dx)# given #2xy+y^2=x+y#?

Answer 1

Elijah Abram

#\frac{1-2y}{2x+2y-1}#

#2xy+y^2 = x+y# #\frac{d}{dx}(2xy+y^{2})=\frac{d}{dx}(x+y)# #2\frac{d}{dx}(xy)+2y\frac{d}{dx}=1+\frac{d}{dx}#

Use product rule and finish it off to get:

#2(x\frac{d}{dx}+y)+2y\frac{d}{dx}=1+\frac{d}{dx}#

to make it clearer, let #\frac{d}{dx}=y'#

so you get: #2(xy'+y)+2yy'=1+y'#

Rearrange to make #y'# the factor to get:

#y'=\frac{1-2y}{2x+2y-1}#

bring back #y'=\frac{d}{dx}#

so you get: #\frac{d}{dx}: \frac{1-2y}{2x+2y-1}#

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Answer 2

Luke Alvarez

#dy/dx=(1-2y)/(2x+2y-1)#

#color(blue)"differentiate implicitly with respect to x"#

#"noting that "d/dx(y)=dy/dx" and "d/dx(y^2)=2ydy/dx#

#"differentiate "xy" using the "color(blue)"product rule"#

#2xdy/dx+2y+2ydy/dx=1+dy/dx#

#dy/dx(2x+2y-1)=1-2y#

#dy/dx=(1-2y)/(2x+2y-1)#

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Answer 3

Cameron Alexander

To find (dy)/(dx) given the equation 2xy + y^2 = x + y, differentiate both sides of the equation with respect to x, then solve for (dy)/(dx). After differentiation and rearrangement, the expression for (dy)/(dx) will be obtained. The process involves applying implicit differentiation and algebraic manipulation to isolate (dy)/(dx) on one side of the equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.