Calculate the second derivative

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Calculate the second derivative

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partial derivatives

implicit function theorem

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#1 h Dec 3, 2021, 6:09 PM

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Find $d^2z$ if $z=f(x,y)$ and $xy + z^2 -zx +zy - 2 = 0$ .

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#2 h Mar 19, 2025, 10:08 AM

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We are interested in computing the total second derivative:

$d^{2}z = dx^{2}\frac{\partial^{2}z}{\partial x^{2}} + dy^{2}\frac{\partial^{2} z}{\partial y^{2}} +2dx dy\frac{\partial^{2} z}{\partial x \partial y}$ (i).

If $z = \frac{(x-y) \pm \sqrt{(x-y)^2-4(1)(xy-2)}}{2} = \frac{x-y}{2} \pm \frac{\sqrt{x^2-6xy+y^2+8}}{2}$ , then we obtain:

$\frac{\partial^{2}z}{\partial x^{2}} = \mp4\left[\frac{y^2-1}{(x^2-6xy+y^2+8)^{3/2}}\right]$ ;

$\frac{\partial^{2}z}{\partial y^{2}} = \mp4\left[\frac{x^2-1}{(x^2-6xy+y^2+8)^{3/2}}\right]$ ;

$\frac{\partial^{2}z}{\partial x \partial y} = \mp4\left[\frac{3-xy}{(x^2-6xy+y^2+8)^{3/2}}\right]$ .

Altogether:

$\textcolor{red}{d^{2}z = \mp\frac{4}{(x^2-6xy+y^2+8)^{3/2}} \cdot \left[(y^2-1)dx^{2} + (x^2-1)dy^{2} + 2(3-xy)dx dy\right]}$ .

This post has been edited 3 times. Last edited by Mathzeus1024, Mar 19, 2025, 10:28 AM

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