If x^1/2 +y^1/2= a^1/2 ,then what is dy/dx?

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Answer 1 · 2018-07-21T16:34:28

#dy/dx=-\sqrt{y/x}#

Given that

#x^{1/2}+y^{1/2}=a^{1/2}#

differentiating above equation w.r.t. #x# on both sides as follows

#d/dx(x^{1/2}+y^{1/2})=d/dx(a^{1/2})#

#1/2x^{1/2-1}+1/2y^{1/2-1}dy/dx=0#

#x^{-1/2}+y^{-1/2}dy/dx=0#

#1/{\sqrtx}+1/{\sqrty}dy/dx=0#

#1/{\sqrty}dy/dx=-1/\sqrtx#

#dy/dx=-\sqrt{y/x}#

Answer 2 · 2018-07-21T16:38:57

#(dy)/(dx)=-sqrt(y/x)#

Here ,

#(1)x^(1/2) +y^(1/2)= a^(1/2)#

Diff. w.r.t. #x# ,we get

#1/2x^(1/2-1)+1/2y^(1/2-1)*(dy)/(dx)=0#

#=>x^(-1/2)+y^(-1/2)(dy)/(dx)=0#

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

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

#=>(dy)/(dx)=-sqrty/sqrtx#

#=>(dy)/(dx)=-sqrt(y/x)#