Answer:
[tex]f_x(8,0)= -\dfrac{8}{ \sqrt{ 473} }[/tex] and [tex]f_y(8,0)= 0[/tex]
Step-by-step explanation:
Given that:
[tex]f(x,y) = \sqrt{537 -x^2 -4y^2}[/tex]
The objective is to determine [tex]f_x(1,0) \ and \ f_y(1,0)[/tex]
Let determine the partial derivative of the function f with respect to x and y, differentiate partially with respect to x. we have:
[tex]\dfrac{\partial f(x,y)}{\partial x}=\dfrac{\partial}{\partial x } ( \sqrt{ 537-x^2-4y^2} )[/tex]
[tex]f_x(x,y)= \dfrac{1}{2 \sqrt{ 537-x^2-4y^2} }\ \dfrac{\partial}{\partial x } ( 537-x^2-4y^2 )[/tex]
[tex]f_x(x,y)= \dfrac{1}{2 \sqrt{ 537-x^2-4y^2} }\ (-2x)[/tex]
[tex]f_x(x,y)= -\dfrac{x}{ \sqrt{ 537-x^2-4y^2} }[/tex]
Thus, [tex]f_x(x,y)= -\dfrac{x}{ \sqrt{ 537-x^2-4y^2} }[/tex]
Differentiate partially with respect to y, we have:
[tex]\dfrac{\partial f(x,y)}{\partial x}=\dfrac{\partial}{\partial x } ( \sqrt{ 537-x^2-4y^2} )[/tex]
[tex]f_y(x,y)= \dfrac{1}{2 \sqrt{ 537-x^2-4y^2} }\ \dfrac{\partial}{\partial x } ( 537-x^2-4y^2 )[/tex]
[tex]f_y(x,y)= \dfrac{1}{2 \sqrt{ 537-x^2-4y^2} }\ (-8y)[/tex]
[tex]f_y(x,y)= -\dfrac{4y}{ \sqrt{ 537-x^2-4y^2} }[/tex]
Thus, [tex]f_y(x,y)= -\dfrac{4y}{ \sqrt{ 537-x^2-4y^2} }[/tex]
Now, substitute 8 for x and o for y into the function [tex]f_x(x,y)= -\dfrac{x}{ \sqrt{ 537-x^2-4y^2} }[/tex]
[tex]f_x(8,0)= -\dfrac{8}{ \sqrt{ 537-8^2-4(0)^2} }[/tex]
[tex]f_x(8,0)= -\dfrac{8}{ \sqrt{ 537-64-0} }[/tex]
[tex]f_x(8,0)= -\dfrac{8}{ \sqrt{ 473} }[/tex]
Also, substitute 8 for x and 0 for y into the function [tex]f_y(x,y)= -\dfrac{4y}{ \sqrt{ 537-x^2-4y^2} }[/tex]
[tex]f_y(8,0)= -\dfrac{4(0)}{ \sqrt{ 537-(8)^2-4(0)^2} }[/tex]
[tex]f_y(8,0)= -\dfrac{0}{ \sqrt{ 537-64-0} }[/tex]
[tex]f_y(8,0)= 0[/tex]
Therefore, [tex]f_x(8,0)= -\dfrac{8}{ \sqrt{ 473} }[/tex] and [tex]f_y(8,0)= 0[/tex]
