Question

Find dy/dy in terms of x and y

Answer 1

`(dy)/(dy)=1`, so I assume you mean "find
`(dy)/(dx)`".

We can rewrite this as an implicit equation to avoid using too much of the chain rule, namely

`y = \sqrt[3]{(e^x (x+1))/(x^2+1)} \implies (x^2+1) y^3 = e^x (x+1)`

Differentiate both sides with respect to
`x` using the product and chain rules.

`2x y^3 + 3(x^2+1) y^2 (dy)/(dx) = e^x(x+1) + e^x`

`\implies 3(x^2+1) y^2 (dy)/(dx) = e^x (x+2) - 2x y^3`

`\implies (dy)/(dx) = (e^x (x+2) - 2x y^3)/(3(x^2+1) y^2)`

Now substitute the original expression for
`y`.

`(dy)/(dx) = \frac{e^x (x+2) - 2x \left(\sqrt[3]{(e^x(x+1))/(x^2+1)}\right)^3}{3(x^2+1) \left(\sqrt[3]{(e^x(x+1))/(x^2+1)}\right)^2}`

`\implies (dy)/(dx) = (e^x (x+2) - (2e^x(x^2+x))/(x^2+1))/(3(x^2+1) \left((e^x(x+1))/(x^2+1)\right)^(2/3))`

`\implies (dy)/(dx) = e^x (x^3-x+2)/(3(x^2+1)^2 (e^(2x/3)(x+1)^(2/3))/((x^2+1)^(2/3)))}`

`\implies (dy)/(dx) = e^(x/3) (x^3-x+2)/(3(x^2+1)^(4/3) (x+1)^(2/3))`

Now, since

`y = \sqrt[3]{(e^x (x+1))/(x^2+1)} = (e^(x/3) (x+1)^(1/3))/((x^2+1)^(1/3))`

we can write

`(dy)/(dx) = e^(x/3) (x^3-x+2)/(3(x^2+1)^(4/3) (x+1)^(2/3)) = (e^(x/3) (x+1)^(1/3))/((x^2+1)^(1/3)) * (x^3-x+2)/(3(x^2+1)^(3/3) (x+1)^(3/3))`

`\implies (dy)/(dx) = y (x^3-x+2)/(3(x^2+1)(x+1))`

Focusing on the rational expression in
`x`, we have the partial fraction expansion

`(x^3 - x + 2)/((x^2 + 1) (x+1)) = a + (bx+c)/(x^2+1) + \frac d{x+1}`

where we have the constant term on the right side because both the numerator and denominator have degree 3.

Writing everything with a common denominator and equating the numerators leads to

`x^3 - x + 2 = a (x^2+1) (x+1) + (bx+c)(x+1) + d(x^2+1) \\\\ = ax^3 + (a+b+d)x^2 + (a+b+c)x + a+c+d`

`\implies \begin{cases} a = 1 \\ a+b+d=0 \\ a+b+c = -1 \\ a+c+d=2 \end{cases}`

`\implies a=1, b=-2, c=0, d=1`

`\implies (x^3 - x + 2)/((x^2 + 1) (x+1)) = 1 - (2x)/(x^2+1) + \frac 1{x+1}`

and it follows that

`\boxed{(dy)/(dx) = \frac y3 \left(1 - (2x)/(x^2+1) + \frac1{x+1}\right)}`