Question

How do you differentiate `(xy)/(x+y)=1`?

Answer 1

`(dy)/(dx)=-y^2/x^2`

Explanation:

Using the quotient rule

`d/(dx)(u/v)=(vu'-uv')/v^2`

`d/(dx)((xy)/(x+y)=1)`

`((x+y)(y+xy')-(xy(1+y')))/(x+y)^2=0`

ie.`(x+y)(y+xy')-xy(1+y')=0`

`cancel(xy)+x^2y'+y^2+cancel(xyy')cancel(-xy)-cancel(xyy')=0`

`x^2(dy)/(dx)+y^2=0`

giving

`(dy)/(dx)=-y^2/x^2`

Answer 2

We can also rewrite the function from the outset to avoid fractions:

`xy=x+y`

Then we see that the right-hand side will use the quotient rule: `d/dx(uv)=u'v+uv'`. Recall that differentiating anything with `y` will cause `dy/dx` to spit out thanks to the chain rule.

Differentiating gives:

`[d/dxx]y+x[d/dxy]=[d/dxx]+[d/dxy]`

`y+xdy/dx=1+dy/dx`

Grouping the `dy/dx` terms:

`xdy/dx-dy/dx=1-y`

Factoring:

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

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

Answer 3

`dy/dx=-1/(x-1)^2`.

Explanation:

We have, `(xy)/(x+y)=1 rArr xy=x+y rArr xy-y=x`

`:. y(x-1)=x rArr y=x/(x-1), xne1`

`:. y'={(x-1)(x)'-(x)(x-1)'}/(x-1)^2............[because, "the Quotient Rule]"`

`={(x-1)(1)-(x)(1-0)}/(x-1)^2=-1/(x-1)^2`

`"Therefore, "dy/dx=-1/(x-1)^2`.