Here's the question you clicked on:
hawkfalcon
if dy/dx =x^2 y^2 then d^2y/dx^2=?
So \[\frac{ dy }{dx }(x^2y^2)\]
Which I get to be -2x^2y/2xy^2 but thats not an option...
\[ \Large \begin{array}{rcl} \frac{dx}{dx} &=&x^2 y^2 \\ \frac{dx}{dx} \left(\frac{dx}{dx}\right) &=& \frac{dx}{dx} (x^2 y^2) \\ \frac{d^2y}{dx^2} &=& (2x)y^2 + x^2\left (2y\frac{dy}{dx} \right) \\ \frac{d^2y}{dx^2} &=& 2xy^2 + 2x^2y(x^2y^2 ) \\ \frac{d^2y}{dx^2} &=& 2xy^2 + 2x^4y^3 \end{array} \]
Oh so I can replace dy/dx with the original function?
In general, when you have an equation, the left and side and right hand side can always be substituted. The exception is when you're trying to prove the equation is true.