Question

How do you find `(dy)/(dx)` given `-x^3y^2+4=5x^2+3y^3`?

Answer 1

`dy/dx=-(10x +3x^2y^2)/(2yx^3+ 9y^2)`

Explanation:

`-x^3y^2 + 4= 5x^2+3y^3`
`" "`
`rArr-3x^2y^2+2ydy/dxxx(-x^3)=10x + 9y^2xxdy/dx`
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`rArr-3x^2y^2-2yx^3dy/dx=10x + 9y^2xxdy/dx`
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`rArr-2yx^3dy/dx- 9y^2xxdy/dx=10x +3x^2y^2`
`" "`
`rArrdy/dx(-2yx^3- 9y^2)=10x +3x^2y^2`
`" "`
`rArrdy/dx=(10x +3x^2y^2)/(-2yx^3- 9y^2)`
`" "`
`rArrdy/dx=-(10x +3x^2y^2)/(2yx^3+ 9y^2)`