Question

if dy/dx =x^2 y^2 then d^2y/dx^2=?

Answer 1

So \[\frac{ dy }{dx }(x^2y^2)\]

Answer 2

Which I get to be -2x^2y/2xy^2 but thats not an option...

Answer 3

\[ \Large
\begin{array}{rcl}
\frac{dx}{dx} &=&x^2 y^2 \\
\frac{dx}{dx} \left(\frac{dx}{dx}\right) &=& \frac{dx}{dx} (x^2 y^2) \\
\frac{d^2y}{dx^2} &=& (2x)y^2 + x^2\left (2y\frac{dy}{dx} \right) \\
\frac{d^2y}{dx^2} &=& 2xy^2 + 2x^2y(x^2y^2 ) \\
\frac{d^2y}{dx^2} &=& 2xy^2 + 2x^4y^3
\end{array}
\]

Answer 4

Oh so I can replace dy/dx with the original function?

Answer 5

Cool! Thanks!

Answer 6

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.