Question

find all second - order partial derivatives for the function ( r=ln|2x + 3y| ).
( r_{xx}=-\frac{4}{(2x + 3y)^{2}} )
( r_{xy}=)

Explanation:

Step1: Find the first - order partial derivative \(r_x\)

Using the chain rule, if \(r = \ln|u|\) where \(u=2x + 3y\), then \(\frac{\partial r}{\partial x}=\frac{1}{u}\cdot\frac{\partial u}{\partial x}\).
Since \(\frac{\partial u}{\partial x}=2\), we have \(r_x=\frac{2}{2x + 3y}\).

Step2: Find the second - order partial derivative \(r_{xy}\)

Differentiate \(r_x=\frac{2}{2x + 3y}\) with respect to \(y\).
Using the quotient rule \((\frac{f}{g})^\prime=\frac{f^\prime g - fg^\prime}{g^{2}}\), here \(f = 2\), \(f^\prime=0\), \(g=2x + 3y\), \(g^\prime = 3\).
So \(r_{xy}=\frac{0\times(2x + 3y)-2\times3}{(2x + 3y)^{2}}=-\frac{6}{(2x + 3y)^{2}}\).

Answer:

\(r_{xy}=-\frac{6}{(2x + 3y)^{2}}\)