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Chapter 11: Problem 64

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=\ln (x-y) $$

Short Answer

Expert verified
The four second partial derivatives are \(z_{xx} = -\frac{1}{{(x-y)}^2}\), \(z_{yy} = -\frac{1}{{(x-y)}^2}\), \(z_{xy} = \frac{1}{{(x-y)}^2}\), \(z_{yx} = \frac{1}{{(x-y)}^2}\). The second mixed partials are indeed equal.

Step by step solution

01

Find the first partial derivatives

The first partial derivatives of \(z\) w.r.t \(x\) and \(y\) are given by: \[z_x = \frac{\partial}{\partial x} \ln (x-y) = \frac{1}{x-y}\] and \[z_y = \frac{\partial}{\partial y} \ln (x-y) = -\frac{1}{x-y}\]
02

Find the second partial derivatives

Now, find the second partial derivatives: \[z_{xx} = \frac{\partial}{\partial x} z_x = \frac{\partial}{\partial x} \left(\frac{1}{x-y}\right) = -\frac{1}{{(x-y)}^2}\] \[z_{yy} = \frac{\partial}{\partial y} z_y = \frac{\partial}{\partial y} \left(-\frac{1}{x-y}\right) = -\frac{1}{{(x-y)}^2}\] \[z_{xy} = \frac{\partial}{\partial y} z_x = \frac{\partial}{\partial y} \left(\frac{1}{x-y}\right) = \frac{1}{{(x-y)}^2}\] \[z_{yx} = \frac{\partial}{\partial x} z_y = \frac{\partial}{\partial x} \left(-\frac{1}{x-y}\right) = \frac{1}{{(x-y)}^2}\]
03

Verify Clairaut's theorem

By Clairaut's theorem, the mixed partial derivatives \(z_{xy}\) and \(z_{yx}\) should be equal. Indeed, they are both equal to \(\frac{1}{{(x-y)}^2}\), so the theorem holds in this case.

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