find y^n by implicit differentiation......... (root)x + (root)y = 1?

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find y^n by implicit differentiation......... (root)x + (root)y = 1?

2006-11-02 18:13:50 · 3 answers · asked by jyoti b 1 in Science & Mathematics ➔ Mathematics

3 answers

sqrt(x) + sqrt(y) = 1

1/[2 sqrt(x)] + 1/[2 sqrt(y)]*(dy/dx) = 1

get (dy/dx) by itself...

(dy/dx) = 2sqrt(y) - [sqrt(y)/sqrt(x)]

2006-11-02 18:28:21 · answer #1 · answered by collegesniper 2 · 0⤊ 0⤋

skip all the tuff stuff and use this formula dy/dx= -(d sqrtx/ dx)/(d sqrt y/dy)
is the simple way to implicit

2006-11-03 02:32:34 · answer #2 · answered by RichUnclePennybags 4 · 0⤊ 0⤋

Start by taking the derivative of both sides with respect to x:
.5/sqrt(x) *dx/dx +.5/sqrt(y)*dy/dx=0
Since dx/dx =1, .5/sqrt(y)*dy/dx=-.5/sqrt(x)
Simplify further to get....dy/dx=-sqrt(y)/sqrt(x)=-sqr(y/x).

2006-11-03 02:19:13 · answer #3 · answered by bruinfan 7 · 0⤊ 0⤋