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

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

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

hawkfalconBest ResponseYou've already chosen the best response.0
Oh so I can replace dy/dx with the original function?
 one year ago

wioBest ResponseYou've already chosen the best response.2
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.
 one year ago
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