Answer:
Actually, this is equal to 1.
Step-by-step explanation:
Hello, please consider the following.
First of all, we assume x and y different from 0.
[tex]xy=4\\\\y=\dfrac{4}{x}\\\\x=\dfrac{4}{x}\\\\\text{So}\\\\y'(x)=\dfrac{-4}{x^2}\\\\y''(x)=\dfrac{-4*(-2)}{x^3}=\dfrac{8}{x^3}\\\\x''(y)=\dfrac{8}{y^2}\\\\\text{So, we can conclude}[/tex]
[tex]\dfrac{d^2y}{dx^2}\cdot \dfrac{d^2x}{dy^2}=\dfrac{8}{x^3}\cdot\dfrac{8}{y^3}\\\\=\dfrac{64}{(xy)^3}\\\\\text{We replace xy by 4}\\\\=\dfrac{64}{4^3}\\\\=\dfrac{64}{64}\\\\=\large \boxed{\sf \bf \ 1 \ }[/tex]
Thank you
