`sqrt(x + y) = 1 + x^2y^2` Find `(dy/dx)` by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 16 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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mathace | (Level 3) Assistant Educator

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`sqrt(x+y)=1+x^2y^2` 

`dy/dxsqrt(x+y)=dy/dx(1+x^2y^2)` 

`1/2sqrt(x+y)^(-1/2)(1+dy/dx)=x^2(2y)dy/dx+y^2(2x)` 

`1/(2sqrt(x+y))+(dy/dx)/(2sqrt(x+y))=2x^2ydy/dx+2xy^2`

`dy/dx((1)/(2sqrt(x+y))-2x^2y)=2xy^2-(1)/(2sqrt(x+y))`

`(dy/dx)((1-4x^2ysqrt(x+y))/(2sqrt(x+y)))=(4x^2y^2sqrt(x+y)-1)/(2sqrt(x+y))`

`dy/dx=(4x^2y^2sqrt(x+y)-1)/(2sqrt(x+y))xx(2sqrt(x+y))/(1-4x^2ysqrt(x+y))`

` dy/dx=(4x^2y^2sqrt(x+y)-1)/(1-4x^2ysqrt(x+y)) `

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