find the linearization of $f(x,y)=\\frac{x}{x + y}$ at the point $(1,5)$. $l(x,y)=$

Explanation:

Step1: Calculate ( f(1,5) )

Substitute ( x = 1 ) and ( y = 5 ) into ( f(x,y)=\frac{x}{x + y} ). ( f(1,5)=\frac{1}{1 + 5}=\frac{1}{6} )

Step2: Find ( f_x(x,y) )

Use the quotient rule ( \left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}} ), where ( u=x ), ( u' = 1 ), ( v=x + y ), ( v'=1 ). ( f_x(x,y)=\frac{(x + y)\times1-x\times1}{(x + y)^{2}}=\frac{y}{(x + y)^{2}} ) Substitute ( x = 1 ), ( y = 5 ) into ( f_x(x,y) ), ( f_x(1,5)=\frac{5}{(1 + 5)^{2}}=\frac{5}{36} )

Step3: Find ( f_y(x,y) )

Use the quotient rule ( \left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}} ), where ( u=x ), ( u' = 0 ), ( v=x + y ), ( v'=1 ). ( f_y(x,y)=\frac{0\times(x + y)-x\times1}{(x + y)^{2}}=-\frac{x}{(x + y)^{2}} ) Substitute ( x = 1 ), ( y = 5 ) into ( f_y(x,y) ), ( f_y(1,5)=-\frac{1}{(1 + 5)^{2}}=-\frac{1}{36} )

Step4: Use the linearization formula ( L(x,y)=f(a,b)+f_x(a,b)(x - a)+f_y(a,b)(y - b) )

Here ( a = 1 ), ( b = 5 ) ( L(x,y)=\frac{1}{6}+\frac{5}{36}(x - 1)-\frac{1}{36}(y - 5) ) Expand the expression: [ \begin{align*} L(x,y)&=\frac{1}{6}+\frac{5x}{36}-\frac{5}{36}-\frac{y}{36}+\frac{5}{36}\ &=\frac{6}{36}+\frac{5x}{36}-\frac{y}{36}\ &=\frac{5x - y+6}{36} \end{align*} ]

Answer:

( L(x,y)=\frac{5x - y + 6}{36} )