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Use implicit derivative to determine \(dy/dx\).

\(\sqrt{xy} = {1+x^2y}\)

 Nov 12, 2020
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Note that   √(xy)  =  x^(1/2) * y^(1/2)

 

So  we have

 

x^(1/2) * y^(1/2)  =  1  + x^2y

 

Using implicit differentiation

 

(1/2)x^(-1/2)* y^(1/2)  + x^(1/2) * (1/2) y^(-1/2) y'  =  0  + 2xy  + x^2 y'  =

 

y^(1/2) / [ 2x^(1/2) ] +  (x^(1/2) / [2 y ^(1/2) ]) y'  = 2xy + x^2 y'

 

Gather the y' terms on the left and everything else on the right we have that

 

y'  [ x^(1/2)/ [ 2y^(1/2)]  - x^2 ]   =  2xy  - y^(1/2) / [ 2x^(1/2) ]

 

y' =  dy/dx =   (  2xy  - y^(1/2) / [ 2x^(1/2) ] )  /  [ x^(1/2)/ [ 2y^(1/2)]  - x^2 ]

 

cool cool cool

 Nov 12, 2020

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