Answer:
A) [tex]\frac{d^2y}{dx^2}=\frac{2y}{x^2}[/tex]
Step-by-step explanation:
Use implicit differentiation to find dy/dx
[tex]x^2y^2=1\\\\\frac{d}{dx}(x^2y^2)=\frac{d}{dx}(1)\\ \\x^2(\frac{d}{dx}y^2)+(\frac{d}{dx}(x^2))y^2 =0\\\\2yx^2\frac{dy}{dx}+2xy^2=0\\\\2xy(x\frac{dy}{dx}+y)=0\\ \\x\frac{dy}{dx}+y=0\\ \\x\frac{dy}{dx}=-y\\ \\\frac{dy}{dx}=-\frac{y}{x}[/tex]
Determine d²y/dx² using dy/dx
[tex]\frac{dy}{dx}=\frac{-y}{x}\\ \\ \frac{d^2y}{dx^2}=\frac{(x)[\frac{d}{dx}(-y)]-(-y)[\frac{d}{dx}(x)] }{x^2}\\ \\ \frac{d^2y}{dx^2}=\frac{-x\frac{dy}{dx}+y}{x^2}\\ \\ \frac{d^2y}{dx^2}=\frac{-x(\frac{-y}{x})+y}{x^2}\\\\ \frac{d^2y}{dx^2}=\frac{-(-y)+y}{x^2}\\\\ \frac{d^2y}{dx^2}=\frac{y+y}{x^2}\\\\ \frac{d^2y}{dx^2}=\frac{2y}{x^2}[/tex]
Helpful tips
- When doing implicit differentiation, make sure to treat "y" as a constant and write dy/dx next to the y-term because it's differentiated with respect to x
- The product rule is [tex]\frac{d}{dx}[f(x)g(x)]=f(x)[\frac{d}{dx}g(x)]+[\frac{d}{dx}f(x)]g(x)[/tex]
- The quotient rule is [tex]\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f(x)[\frac{d}{dx}g(x)]-g(x)[\frac{d}{dx}f(x)]}{(g(x))^2}[/tex]
- [tex]\frac{dy}{dx}[/tex] represents the first derivative of y with respect to x and [tex]\frac{d^2y}{dx^2}[/tex] represents the second derivative of y with respect to x
Let me know if you have any more questions!
