/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Find the four second partial der... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 65

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=e^{x} \tan y $$

Short Answer

Expert verified
The four second partial derivatives are: \( \frac{{\partial^{2}z}}{{\partial x^2}} = e^{x} \tan(y) \), \( \frac{{\partial^{2}z}}{{\partial y^2}} = 2e^{x} \sec^2(y) \tan(y) \), \( \frac{{\partial^{2}z}}{{\partial x\partial y}} = e^{x} \sec^2(y) \), \( \frac{{\partial^{2}z}}{{\partial y\partial x}} = e^{x} \sec^2(y) \). The second mixed partial derivatives are indeed equal: \( \frac{{\partial^{2}z}}{{\partial y \partial x}} = \frac{{\partial^{2}z}}{{\partial x \partial y}} = e^{x} \sec^2(y) \).

Step by step solution

01

Compute the first partial derivatives with respect to x and y

Firstly, find the first partial derivatives of the function with respect to x and y. The derivative of \( e^{x} \) with respect to x is \( e^{x} \), and the derivative of \( \tan(y) \) with respect to y is \( \sec^2(y) \). Hence, \n\n\[ \frac{{\partial z}}{{\partial x}} = e^{x} \tan(y) \] \n\nand\n\n\[ \frac{{\partial z}}{{\partial y}} = e^{x} \sec^2(y) \]
02

Compute the second partial derivatives

Having the first partial derivatives, it is possible now to calculate the four second partial derivatives. The derivative of \( e^{x} \tan(y) \) with respect to x is \( e^{x} \tan(y) \), and with respect to y is \( e^{x} \sec^2(y) \). This gives \n\n\[ \frac{{\partial^{2}z}}{{\partial x^2}} = e^{x} \tan(y) \] \n\nand \n\n\[ \frac{{\partial^{2}z}}{{\partial x \partial y}} = e^{x} \sec^2(y) \].\n\nThe derivative of \( e^{x} \sec^2(y) \) with respect to x is \( e^{x} \sec^2(y) \), and with respect to y is \( 2e^{x} \sec(y) \tan(y) \sec(y) \). This gives \n\n\[ \frac{{\partial^{2}z}}{{\partial y^2}} = 2e^{x} \sec^2(y) \tan(y) \] \n\nand \n\n\[ \frac{{\partial^{2}z}}{{\partial y \partial x}} = e^{x} \sec^2(y) \].
03

Verify the equality

From steps 1 and 2, it can be observed that the mixed second partial derivatives are equal, that is \n\n\[ \frac{{\partial^{2}z}}{{\partial y \partial x}} = \frac{{\partial^{2}z}}{{\partial x \partial y}} = e^{x} \sec^2(y)\], \n\nthus confirming the Clairaut's theorem which states that if all second partial derivatives of a function are continuous in a neighborhood, then the mixed derivatives are equal in that neighborhood.

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