Jawaban Carilah polinom taylor sampai derajat 2 soal terlampir​

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Pertanyaan

Carilah polinom taylor sampai derajat 2 soal terlampir​

Jawaban #1 untuk Pertanyaan: Carilah polinom taylor sampai derajat 2 soal terlampir​

Jawab:

[tex]\displaystyle f(x,y)=\frac{3x^3}{2-2x^2-2y^2}\\f(1,1)=\frac{3\cdot1^3}{2-2\cdot1^2-2\cdot1^2}\\f(x,y)=\frac{3x^3}{2-2x^2-2y^2}\\f(1,1)=\frac{3}{2-2-2}\\f(1,1)=-\frac{3}{2}[/tex]

[tex]\displaystyle f_x(x,y)=\frac{9x^2(2-2x^2-2y^2)-3x^3(-4x)}{(2-2x^2-2y^2)^2}\\f_x(x,y)=\frac{18x^2-18x^4-18x^2y^2+12x^4}{(2-2x^2-2y^2)^2}\\f_x(x,y)=\frac{(1-y^2)18x^2-6x^4}{(2-2x^2-2y^2)^2}\\f_x(1,1)=\frac{(1-1^2)18\cdot1^2-6\cdot1^4}{(2-2\cdot1^2-2\cdot1^2)^2}\\f_x(1,1)=\frac{0\cdot18-6}{(2-2-2)^2}\\f_x(1,1)=\frac{-6}{4}\\f_x(1,1)=-\frac{3}{2}[/tex]

[tex]\displaystyle f_y(x,y)=\frac{0(2-2x^2-2y^2)-3x^3(-4y)}{(2-2x^2-2y^2)^2}\\f_y(x,y)=\frac{12x^3y}{(2-2x^2-2y^2)^2}\\f_y(1,1)=\frac{12\cdot1^3\cdot1}{(2-2\cdot1^2-2\cdot1^2)^2}\\f_y(1,1)=\frac{12}{(2-2-2)^2}\\f_y(1,1)=\frac{12}{4}\\f_y(1,1)=3[/tex]

[tex]\displaystyle f_{xx}(x,y)=\frac{((1-y^2)36x-24x^3)(2-2x^2-2y^2)^2+((1-y^2)18x^2-6x^4)(8x)}{((2-2x^2-2y^2)^2)^2}\\f_{xx}(x,y)=\frac{(1-y^2)36x-24x^3}{(2-2x^2-2y^2)^2}+\frac{8x((1-y^2)18x^2-6x^4)}{(2-2x^2-2y^2)^4}\\f_{xx}(1,1)=\frac{(1-1^2)36\cdot1-24\cdot1^3}{(2-2\cdot1^2-2\cdot1^2)^2}+\frac{8\cdot1((1-1^2)18\cdot1^2-6\cdot1^4)}{(2-2\cdot1^2-2\cdot1^2)^4}\\f_{xx}(1,1)=\frac{(1-1)36-24}{(2-2-2)^2}+\frac{8((1-1)18-6)}{(2-2-2)^4}\\f_{xx}(1,1)=-\frac{24}{4}-\frac{48}{16}\\f_{xx}(1,1)=-6-3\\f_{xx}(1,1)=-9[/tex]

[tex]\displaystyle f_{xy}(x,y)=\frac{36x^2y(2-2x^2-2y^2)^2-12x^3y(-8x)}{((2-2x^2-2y^2)^2)^2}\\f_{xy}(x,y)=\frac{36x^2y(2-2x^2-2y^2)^2+96x^4y}{(2-2x^2-2y^2)^4}\\f_{xy}(1,1)=\frac{36\cdot1^2\cdot1(2-2\cdot1^2-2\cdot1^2)^2+96\cdot1^4\cdot1}{(2-2\cdot1^2-2\cdot1^2)^4}\\f_{xy}(1,1)=\frac{36(2-2-2)^2+96}{(2-2-2)^4}\\f_{xy}(1,1)=\frac{36\cdot4+96}{16}\\f_{xy}(1,1)=9+6\\f_{xy}(1,1)=15[/tex]

[tex]\displaystyle f_{yy}(x,y)=\frac{12x^3(2-2x^2-2y^2)^2-12x^3y(-8y)}{((2-2x^2-2y^2)^2)^2}\\f_{yy}(x,y)=\frac{12x^3(2-2x^2-2y^2)^2+96x^3y^2}{(2-2x^2-2y^2)^4}\\f_{yy}(1,1)=\frac{12(2-2-2)^2+96}{(2-2-2)^4}\\f_{yy}(1,1)=\frac{48+96}{16}\\f_{yy}(1,1)=3+6\\f_{yy}(1,1)=9[/tex]

[tex]\displaystyle f(x,y)=f(a,b)+\frac1{1!}((x-a)f_x(a,b)+(y-b)f_y(a,b))+\frac1{2!}(f_{xx}(a,b)(x-a)^2+2(x-a)(y-b)f_{xy}(a,b)+f_{yy}(a,b)(y-b)^2)\\f(x,y)=f(1,1)+(x-1)f_x(1,1)+(y-1)f_y(1,1)+\frac1{2}(f_{xx}(1,1)(x-1)^2+2(x-1)(y-1)f_{xy}(1,1)+f_{yy}(1,1)(y-1)^2)\\f(x,y)=-\frac32-\frac32(x-1)+3(y-1)+\frac1{2}(-9(x-1)^2+30(x-1)(y-1)+9(y-1)^2)[/tex]

Beberapa konsep yang dipakai:

[tex]\displaystyle \triangleright~f_x(x,y)=\frac{\partial}{\partial x}f(x,y)\\\triangleright~f_y(x,y)=\frac{\partial}{\partial y}f(x,y)\\\triangleright~f_{xx}(x,y)=\frac{\partial^2}{\partial x^2}f(x,y)\\\triangleright~f_{yy}(x,y)=\frac{\partial^2}{\partial y^2}f(x,y)\\\triangleright~f_{xy}(x,y)=\frac{\partial}{\partial y}(f_x(x,y))\\\triangleright~\frac{d}{dx}(y_1+y_2+\cdots+y_n)=y’_1+y’_2+\cdots+y’_n\\\triangleright~\frac{d}{dx}\left(\frac uv\right)=\frac{u’v-uv’}{v^2}\\\triangleright~\frac{d}{dx}(ax^n)=anx^{n-1}[/tex]

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