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

find all second - order partial derivatives for the function ( r=ln|2x + 3y| ).\n( r_{xx}=-\frac{4}{(2x + 3y)^{2}} )\n( r_{xy}=-\frac{6}{(2x + 3y)^{2}} )\n( r_{yx}= )

find all second - order partial derivatives for the function ( r=ln|2x + 3y| ).\n( r_{xx}=-\frac{4}{(2x + 3y)^{2}} )\n( r_{xy}=-\frac{6}{(2x + 3y)^{2}} )\n( r_{yx}= )

Answer

Explanation:

Step1: Find the first - order partial derivative (r_y)

Use the chain rule. If (r = \ln|2x + 3y|), then (r_y=\frac{3}{2x + 3y}) (since (\frac{d}{du}\ln|u|=\frac{1}{u}) and (u = 2x+3y), (\frac{\partial u}{\partial y}=3)).

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

Differentiate (r_y=\frac{3}{2x + 3y}) with respect to (x). Using the quotient rule (\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}), where (u = 3), (u'=0), (v=2x + 3y), (v'=2). [ \begin{align*} r_{yx}&=\frac{0\times(2x + 3y)-3\times2}{(2x + 3y)^{2}}\ &=\frac{- 6}{(2x + 3y)^{2}} \end{align*} ]

Answer:

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