How do you find #dy/dx# by implicit differentiation given #y=1/(x+y)#?

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Answer 1 · 2017-09-11T17:29:27

# dy/dx=-y^2/(y^2+1)," or, equivalently, "dy/dx=-1/{1+(x+y)^2}.#

We rewrite the given eqn. #y=1/(x+y)," as, "y(x+y)=1,#

# i.e., xy+y^2=1.#

#:. xy=1-y^2, or, x=(1-y^2)/y.#

#:. x=1/y-y^2/y=1/y-y.#

Diff.ing both sides, w.r.t. #y,# we have,

#dx/dy=d/dy(1/y-y)=-1/y^2-1=-(1+y^2)/y^2.#

#rArr dy/dx=1/(dx/dy)=-y^2/(y^2+1).#

Equivalently, this can be written as,

#dy/dx=-y^2/(y^2+1)=-y^2/{y^2(1+1/y^2),#

#=-1/(1+1/y^2),# &, since, #1/y=(x+y),#

# dy/dx=-1/{1+(x+y)^2},#

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