How do you find #dy/dx# by implicit differentiation given #sqrtx+sqrty=x+y#?

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Answer 1 · 2018-02-06T08:33:51

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

We have #x^(1/2)+y^(1/2)=x+y#

First we take #d/dx# of each term:
#d/dx[x^(1/2)]+d/dx[y^(1/2)]=d/dx[x]+d/dx[y]#

#x^(-1/2)/2+d/dx[y^(1/2)]=1+d/dx[y]#

Using the chain rule: #d/dx=dy/dx*d/dy#

#x^(-1/2)/2+dy/dxd/dy[y^(1/2)]=1+dy/dxd/dy[y]#

#x^(-1/2)/2+dy/dxy^(-1/2)/2=1+dy/dx1#

#dy/dxy^(-1/2)/2-dy/dx1=1-x^(-1/2)/2#

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

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

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

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

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

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

Answer 2 · 2018-02-06T08:34:40

#(1/(2sqrtx)-1)/(1-1/(2sqrty))#

Differentiate the two sides:

#(dsqrtx)/(dx)+(dsqrty)/(dx)=dx/dx+dy/dx#

#1/(2sqrtx)+1/(2sqrty)*dy/dx=1+dy/dx#

Leave #dy/dx# alone:

#dy/dx-1/(2sqrty)*dy/dx=1/(2sqrtx)-1#

#dy/dx(1-1/(2sqrty))=1/(2sqrtx)-1#

#dy/dx=(1/(2sqrtx)-1)/(1-1/(2sqrty))#