Find the third Taylor polynomial of the function f(x,y)= =1+5xy+3²y at (1,2)
Given f(x,y)=1−5xy+32y=1−5xy+9yf(x,y)=1-5xy+3^2y=1-5xy+9yf(x,y)=1−5xy+32y=1−5xy+9y
f(1,2)=1−5(1)(2)+9(2)=1−10+18=9f(1,2)=1-5(1)(2)+9(2)=1-10+18=9f(1,2)=1−5(1)(2)+9(2)=1−10+18=9
fx=−5yfx(1,2)=−10fy=−5x+9fy(1,2)=−5(1)+9=4fxx=0fyy=0f_x=-5y\\ f_x(1,2)=-10 \\ f_y=-5x+9 \\ f_y(1,2)=-5(1)+9=4 \\ f_{xx}=0\\ f_{yy}=0 \\fx=−5yfx(1,2)=−10fy=−5x+9fy(1,2)=−5(1)+9=4fxx=0fyy=0
f(x,y)=f(a,b)+(x−1)fx(1,2)+(y−2)fy+fxx×(x−1)22!+....f(x,y)=f(a,b)+(x-1)f_x(1,2)+(y-2)f_y+\dfrac{f_{xx}\times (x-1)^2}{2!}+....f(x,y)=f(a,b)+(x−1)fx(1,2)+(y−2)fy+2!fxx×(x−1)2+....
=9+(x−1)×−10+(y−2)×4+0+...=9−10x+10+4y−8=4y−10x+11=9+(x-1)\times -10+(y-2)\times 4+0+...\\[9pt]=9-10x+10+4y-8\\[9pt]=4y-10x+11=9+(x−1)×−10+(y−2)×4+0+...=9−10x+10+4y−8=4y−10x+11
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