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hawkfalcon

  • 2 years ago

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

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  1. hawkfalcon
    • 2 years ago
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    So \[\frac{ dy }{dx }(x^2y^2)\]

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

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

  4. hawkfalcon
    • 2 years ago
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    Oh so I can replace dy/dx with the original function?

  5. hawkfalcon
    • 2 years ago
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    Cool! Thanks!

  6. wio
    • 2 years ago
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    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.

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