Question

If $y=\frac{1}{x},$ then the value of $\frac{dy}{\sqrt{1+y^4}}+\frac{dx}{\sqrt{1+x^4}}+1$ is equal to

Answer 1

$y=\frac{1}{x}\Rightarrow \frac{dy}{dx}=\frac{-1}{x^2}\Rightarrow x^{2}dy+dx=0$

$\Rightarrow \frac{x^2}{\sqrt{1+x^4}}dy+\frac{dx}{\sqrt{1+x^4}}=0$

$\Rightarrow \frac{dy}{\sqrt{\frac{1}{x^4}+1}}+\frac{dx}{\sqrt{1+x^4}}=0$

$\Rightarrow \frac{dy}{\sqrt{1+y^4}}+\frac{dx}{\sqrt{1+x^4}}=0$

Hence, $\frac{dy}{\sqrt{1+y^4}}+\frac{dx}{\sqrt{1+x^4}}+1=1$