If x root (1+y) + y root (1+x) = 0 Prove that dy/dx = -1/(1+x)2 experts do help!!
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x √(1+y) + y √(1+x) = 0
x √(1+y) = - y √(1+x)
square on both sides:
x² (1+y) = y² (1+x)
y² (1+x) - x² y - x² = 0
y = [ x² + √(x⁴+4x+4x³) ] / [2(1+x)]
= [x² + x(x+2) ] /[2(1+x)]
y = x or -x / (1+x)
or, y = x or -1 + 1/(1+x)
Differentiate wrt x:
dy/dx = 1 or -1/(1+x)²
x √(1+y) = - y √(1+x)
square on both sides:
x² (1+y) = y² (1+x)
y² (1+x) - x² y - x² = 0
y = [ x² + √(x⁴+4x+4x³) ] / [2(1+x)]
= [x² + x(x+2) ] /[2(1+x)]
y = x or -x / (1+x)
or, y = x or -1 + 1/(1+x)
Differentiate wrt x:
dy/dx = 1 or -1/(1+x)²
kvnmurty:
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