Will you Differentiate and add

Given that $y = \dfrac 1x$ $\implies \dfrac {dy}{dx} = -\dfrac 1{x^2}$ $\implies x^2 \ dy = -dx$ .

Then we have:

$\begin{aligned} \dfrac {dy}{\sqrt{1+y^4}} + \dfrac {dx}{\sqrt{1+x^4}} & = \dfrac {dy}{\sqrt{1+\frac 1{x^4}}} + \dfrac {dx}{\sqrt{1+x^4}} \\ & = \dfrac {x^2 dy}{\sqrt{x^4+1}} + \dfrac {dx}{\sqrt{1+x^4}} \\ & = \dfrac {- dx}{\sqrt{1+x^4}} + \dfrac {dx}{\sqrt{1+x^4}} \\ & = \boxed 0 \end{aligned}$

If $y = \frac{1}{x}$ , then $dy = -\frac{1}{x^2} dx$ . Substituting both these values into the above expression gives:

$\frac{dy}{\sqrt{1 + y^4}} + \frac{dx}{\sqrt{1+x^4}} = \frac{\frac{-1}{x^2}dx}{\sqrt{1 + \frac{1}{x^4}}} + \frac{dx}{\sqrt{1+x^4}};$

or $-\frac{1}{x^2} \cdot \frac {x^2 dx}{\sqrt{1+x^4}} + \frac{dx}{\sqrt{1+x^4}};$

or $\frac{dx}{\sqrt{1+x^4}} - \frac{dx}{\sqrt{1+x^4}};$

or $\boxed{0}.$