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Math Help - ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η) second derivative problematic

  1. #1
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    ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η) second derivative problematic

    Hello guys, I'm pretty new here and I have a problem. I couldn't take 2nd derivative of this equation;

    ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η)

    ∂^2/(∂^2 y)=∂/∂y (ξ_y  ∂/∂ξ+η_y  ∂/∂η) second derivative problematic-eq.jpg

    Thank you so much for your helps in advance
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  2. #2
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    Re: ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η) second derivative problematic

    That's not an equation- you are just saying that those two operators are the same. But I don't understand what it is that you want to take the second derivative of. The second derivative of the left side would be \frac{\partial^4}{\partial y^4} and the second derivative of the right \frac{\partial^3}{\partial y^3}\left(\xi_y\frac{\partial}{\partial \xi}+ \eta_y\frac{\partial}{\partial \eta}\right)
    but I doubt that is what you are asking.
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    Re: ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η) second derivative problematic

    ∂^2/(∂^2 y)=∂/∂y (ξ_y  ∂/∂ξ+η_y  ∂/∂η) second derivative problematic-eq2.jpg

    I couldn't expand the right side of the equation by taking 2nd derivative of it. Is it a bit more clear now?
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  4. #4
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    Re: ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η) second derivative problematic

    Quote Originally Posted by gorkyerdogan View Post
    Click image for larger version. 

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    I couldn't expand the right side of the equation by taking 2nd derivative of it. Is it a bit more clear now?
    just use the product rule for derivatives

    \frac{\partial}{\partial y}\left(\xi_y \frac{\partial}{\partial \xi}+\eta_y \frac{\partial}{\partial \eta}\right)=\frac{\partial \xi_y}{\partial y}\frac{\partial}{\partial \xi}+\xi_y \frac{\partial}{\partial y \partial \xi}+\frac{\partial \eta_y}{\partial y}\frac{\partial}{\partial \eta}+\eta_y \frac{\partial }{\partial y \partial \eta}

    I hope you understand that these are operators.
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    Re: ∂^2/(∂^2 y)=∂/∂y (ξ_y ∂/∂ξ+η_y ∂/∂η) second derivative problematic

    Thank you very much mate. This is so helpful for me.
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