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

2 Answers
Feb 6, 2018

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

Explanation:

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)

Feb 6, 2018

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

Explanation:

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))