Let $f ( x , y ) = \ln \left( \sqrt { x ^ { 2 } + y ^ { 2 } } \right)$ . Then $f _ { x y } ( x , y )$ is

A) $- \frac { 2 x y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
B) $\frac { 2 x y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
C) $- \frac { x ^ { 2 } y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
D) $\frac { x ^ { 2 } y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
E) $- \frac { y ^ { 2 } x } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$

Question

Let $f ( x , y ) = \ln \left( \sqrt { x ^ { 2 } + y ^ { 2 } } \right)$ . Then $f _ { x y } ( x , y )$ is

A) $- \frac { 2 x y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
B) $\frac { 2 x y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
C) $- \frac { x ^ { 2 } y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
D) $\frac { x ^ { 2 } y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$
E) $- \frac { y ^ { 2 } x } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } }$

Quizplus · Accepted Answer

To find $f_{xy}(x, y)$, we first differentiate $f(x, y)$ with respect to $x$ and then differentiate the result with respect to $y$. First, differentiate $f(x, y) = \ln(\sqrt{x^2 + y^2})$ with respect to $x$:$$f_x(x, y) = \frac{1}{\sqrt{x^2 + y^2}} \cdot \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x = \frac{x}{x^2 + y^2}$$Then, differentiate $f_x(x, y)$ with respect to $y$:$$f_{xy}(x, y) = \frac{\partial}{\partial y} \left(\frac{x}{x^2 + y^2}\right) = x \cdot \frac{\partial}{\partial y} \left(\frac{1}{x^2 + y^2}\right) = x \cdot \left(-\frac{2y}{(x^2 + y^2)^2}\right) = -\frac{2xy}{(x^2 + y^2)^2}$$Thus, the correct answer is $-\frac{2xy}{(x^2 + y^2)^2}$, which matches choice A.

Let f(x,y)=ln⁡(x2+y2)f ( x , y ) = \ln \left( \sqrt { x ^ { 2 } + y ^ { 2 } } \right)f(x,y)=ln(x2+y2​)

Let $f ( x , y ) = \ln \left( \sqrt { x ^ { 2 } + y ^ { 2 } } \right)$