How do you find #dy/dx# given #x=y+2sqrty#?

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Answer 1 · 2017-01-23T10:25:16

#(dy)/(dx) = 1 - 1/sqrt(x+1)#

We have:

#x= y+2sqrt(y)#

Completing the square at the second member:

#x+1 = y+2sqrt(y)+1 = (1+sqrt(y))^2#

we can extract the square root, and as #1+sqrt(y) > 0# we keep only the positive value of the root at the first member:

#sqrt(x+1) = 1+sqrt(y)#

#sqrt(y) = sqrt(x+1) -1#

#y = (sqrt(x+1) -1)^2 = (x+1) -2sqrt(x+1) +1 = x -2sqrt(x+1) +2#

we have thus made #y(x)# explicit, so that:

#(dy)/(dx) = 1 - 1/sqrt(x+1)#

Answer 2 · 2017-01-23T11:40:19

#x>=-1 and y >=0#. See the Socratic graph.
#y'=sqrty/(sqrty+ 1)#.

graph{y+sqrty-x=0 [-10, 10, -5, 5]}

For real x, #y >= 0#

#(dx)/(dy)=1+1/sqrty#

#y'=1/((dx)/(dy))=sqrty/(sqrty+ 1)#.

Of course, inversely,

#y = sqrt(1+x) -1>=0#, giving #x >=0#