if x(1+x)^1/2+y(1+x)^1/2=0 then (1+x)^2dy/dx
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Hey mate,
Here is your answer :
Step-by-step explanation:
given
x√(1+y)+y√(1+x)=0
x√(1+y) = - y√(1+x)
squareing on both sides
x²(1+y) = - y²(1+x)
x²+x²y = -y²- y²x
x²-y² = y²x-x²y
(x+y)(x-y)=-xy(x-y)
x+y=-xy
case1
x= -y(x+1)---------i
case2
y=-x/(X+1)---------i
according to problem
dy= (x+1) 2−1(x+1)−1(−x)
on solving.
dx
= (x+1) 2−1(x+1) 2
dy
dx
=−1
dy
on solving ...
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