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

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Chapter 2, 2.5 - Problem 8 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`sqrt(xy)=x^2y+1`

Differentiating both sides with respect to x,

`1/2(xy)^(-1/2)d/dx(xy)=x^2d/dx(y)+yd/dx(x^2)`

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

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

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

`((x-2x^2sqrt(xy))/(2sqrt(xy)))dy/dx=(4xysqrt(xy)-y)/(2sqrt(xy))`

`dy/dx=(4xysqrt(xy)-y)/(x-2x^2sqrt(xy))`

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