Evaluate $f _ { x }$ and $f _ { y }$ for the function $f ( x , y ) = \frac { 7 x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } }$ at the point $( 4,8 )$ . Round your answer to two decimal places.
A) $f _ { x } ( 4,8 ) =$ 5.01 and $f _ { y } ( 4,8 ) =$ 0.63
B) $f _ { x } ( 4,8 ) =$ 4.51 and $f _ { y } ( 4,8 ) =$ 0.63
C) $f _ { x } ( 4,8 ) =$ 3.13 and $f _ { y } ( 4,8 ) =$ 6.26
D) $f _ { x } ( 4,8 ) =$ 5.01 and $f _ { y } ( 4,8 ) =$ 0.13
E) $f _ { x } ( 4,8 ) =$ 3.13 and $f _ { y } ( 4,8 ) =$ 6.26.

Question

Quizplus · Accepted Answer

To find $f_x$ and $f_y$, we differentiate $f(x, y) = \frac{7xy}{\sqrt{x^2 + y^2}}$ with respect to $x$ and $y$ respectively. For $f_x$, we use the quotient rule and the chain rule:$$f_x = \frac{7y(x^2 + y^2)^{1/2} - 7xy \cdot \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2x}{x^2 + y^2}$$Plugging in $x = 4$ and $y = 8$, we simplify to get $f_x(4,8) = 5.01$.For $f_y$, similarly:$$f_y = \frac{7x(x^2 + y^2)^{1/2} - 7xy \cdot \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot 2y}{x^2 + y^2}$$Plugging in $x = 4$ and $y = 8$, we simplify to get $f_y(4,8) = 0.63$.Therefore, the correct answer is $f_x(4,8) = 5.01$ and $f_y(4,8) = 0.63$.

Evaluate fxf _ { x }fx​ And fyf _ { y }fy​

Evaluate $f _ { x }$ And $f _ { y }$