How do you differentiate #x+y=xy#?

Answer 1

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

#"differentiate "color(blue)"implicitly with respect to x"#
#"differentiate " xy" using the "color(blue)"product rule"#
#rArr1+dy/dx=xdy/dx+y#
#rArrdy/dx(1-x)=y-1#
#rArrdy/dx=(y-1)/(1-x)#
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Answer 2

To differentiate ( x + y = xy ), we will first rewrite the equation to solve for ( y ) in terms of ( x ). Then, we will use implicit differentiation to find the derivatives with respect to ( x ).

First, rewrite the equation as ( y = xy - x ).

Now, differentiate both sides of the equation with respect to ( x ), using the product rule on the right side:

[ \frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y - 1 ]

Now, solve for ( \frac{{dy}}{{dx}} ):

[ \frac{{dy}}{{dx}} - x\frac{{dy}}{{dx}} = y - 1 \ \frac{{dy}}{{dx}}(1 - x) = y - 1 \ \frac{{dy}}{{dx}} = \frac{{y - 1}}{{1 - x}} ]

Next, substitute the expression for ( y ) from the original equation:

[ \frac{{dy}}{{dx}} = \frac{{xy - x - 1}}{{1 - x}} ]

This is the derivative of ( y ) with respect to ( x ) for the given equation ( x + y = xy ).

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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.

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.

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.

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.

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