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

3 Answers
Jan 20, 2017

#(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#

Jan 22, 2017

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)#

Jan 22, 2017

# 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#.