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hawkfalcon
 2 years ago
Best ResponseYou've already chosen the best response.0So \[\frac{ dy }{dx }(x^2y^2)\]

hawkfalcon
 2 years ago
Best ResponseYou've already chosen the best response.0Which I get to be 2x^2y/2xy^2 but thats not an option...

wio
 2 years ago
Best ResponseYou've already chosen the best response.2\[ \Large \begin{array}{rcl} \frac{dx}{dx} &=&x^2 y^2 \\ \frac{dx}{dx} \left(\frac{dx}{dx}\right) &=& \frac{dx}{dx} (x^2 y^2) \\ \frac{d^2y}{dx^2} &=& (2x)y^2 + x^2\left (2y\frac{dy}{dx} \right) \\ \frac{d^2y}{dx^2} &=& 2xy^2 + 2x^2y(x^2y^2 ) \\ \frac{d^2y}{dx^2} &=& 2xy^2 + 2x^4y^3 \end{array} \]

hawkfalcon
 2 years ago
Best ResponseYou've already chosen the best response.0Oh so I can replace dy/dx with the original function?

wio
 2 years ago
Best ResponseYou've already chosen the best response.2In 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.
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