Question #339592

Find the first partial derivative of

f(x,y) = (4xy)/(4x+y)(4x-y)/(4x+y) at the point (4,2) for both x and y.



1
Expert's answer
2022-05-13T03:03:14-0400

f(x,y)=(4xy)/(4x+y)f(x,y)=(4x-y)/(4x+y)

fx=(4xy)x(4x+y)(4x+y)x(4xy)(4x+y)2=4(4x+y)4(4xy)(4x+y)2=8y(4x+y)2f_x=\frac{(4x-y)_x'(4x+y)-(4x+y)_x'(4x-y)}{(4x+y)^2}=\frac{4(4x+y)-4(4x-y)}{(4x+y)^2}=\frac{8y}{(4x+y)^2}

At the point (4,2)

fx=8(2)(4(4)+2)2=16/324=4/81f_x=\frac{8(2)}{(4(4)+2)^2}=16/324=4/81

fy=(4xy)y(4x+y)(4x+y)y(4xy)(4x+y)2=1(4x+y)1(4xy)(4x+y)2=8x(4x+y)2f_y=\frac{(4x-y)_y'(4x+y)-(4x+y)_y'(4x-y)}{(4x+y)^2}=\frac{-1(4x+y)-1(4x-y)}{(4x+y)^2}=\frac{-8x}{(4x+y)^2}

At the point (4,2)

fy=8(4)(4(4)+2)2=32/324=8/81f_y=\frac{-8(4)}{(4(4)+2)^2}=-32/324=-8/81


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