Question

Find dy/dy in terms of x and y

Answer 1

$\frac{dy}{dy}=1$, so I assume you mean "find $\frac{dy}{dx}$".

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

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

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

$2x y^3 + 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x(x+1) + e^x$

$\implies 3(x^2+1) y^2 \dfrac{dy}{dx} = e^x (x+2) - 2x y^3$

$\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - 2x y^3}{3(x^2+1) y^2}$

Now substitute the original expression for $y$.

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

$\implies \dfrac{dy}{dx} = \dfrac{e^x (x+2) - \frac{2e^x(x^2+x)}{x^2+1}}{3(x^2+1) \left(\frac{e^x(x+1)}{x^2+1}\right)^{2/3}}$

$\implies \dfrac{dy}{dx} = e^x \dfrac{x^3-x+2}{3(x^2+1)^2 \frac{e^{2x/3}(x+1)^{2/3}}{(x^2+1)^{2/3}}}}$

$\implies \dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}}$

Now, since

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

we can write

$\dfrac{dy}{dx} = e^{x/3} \dfrac{x^3-x+2}{3(x^2+1)^{4/3} (x+1)^{2/3}} = \dfrac{e^{x/3} (x+1)^{1/3}}{(x^2+1)^{1/3}} \times \dfrac{x^3-x+2}{3(x^2+1)^{3/3} (x+1)^{3/3}}$

$\implies \dfrac{dy}{dx} = y \dfrac{x^3-x+2}{3(x^2+1)(x+1)}$

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

$\dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = a + \dfrac{bx+c}{x^2+1} + \dfrac d{x+1}$

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

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

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

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

$\implies \dfrac{x^3 - x + 2}{(x^2 + 1) (x+1)} = 1 - \dfrac{2x}{x^2+1} + \dfrac 1{x+1}$

and it follows that

$\boxed{\dfrac{dy}{dx} = \dfrac y3 \left(1 - \dfrac{2x}{x^2+1} + \dfrac1{x+1}\right)}$