# Use implicit differentiation to find dy/dx given the relation (y^5+x^2)y^3 = 1 + ye^(x^2)

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### 2 Answers

We have

`y^3(y^5+x^2) = 1 + ye^(x^2)`

Differentiating both sides with respect to `x` we get

`3y^2((dy)/(dx))(y^5 + x^2) + y^3(5y^4((dy)/(dx))+ 2x) `

`= 0 + 2xye^(x^2) + ((dy)/(dx))e^(x^2)`

[Remember that `d/(dx)f(y) = f'(y)((dy)/(dx))` ]

Writing `y' =(dy)/(dx)` this gives

`y'[3y^2(y^5+x^2) + 5y^7] + 2xy^3 = 2xye^(x^2) + y'e^(x^2)`

`implies`

`y'(8y^7+ 3x^2y^2 - e^(x^2)) = 2xy(e^(x^2)-y^2)`

`implies`

`y' = (dy)/(dx) = (2xy(e^(x^2)-y^2))/(8y^7 + 3x^2y^2 - e^(x^2))`

**answer**

`(y^5+x^2)y^3=1+ye^(x^2)`

Before to use derivative, we se togheter all y monoms:

`y^8+x^2y^3-ye^(x^2)=1```

`d/dx(y^8+x^2y^3-ye^(x^2))=d/dx 1`

`8y^7 y' +2xy^3+3x^2 y^2 y'-y'e^(x^2)-2xy e^(x^2)=0`

`y'(8y^7+3x^2y^2-e^(x^2))=2xy(e^(x^2)-y^2)`

`y'=(2xy(e^(x^2)-y^2))/(8y^7+3x^2y^2-e^(x^2))`

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