The expansion of f(x,y), is

Question

TuteeHUB · Accepted Answer

f(0,0)+$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…$

TuteeHUB · Answer

f(0,0)+$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+….$

TuteeHUB · Answer

f(0,0)+$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…$

TuteeHUB · Answer

f(0,0)+$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…$

TuteeHUB · Answer

f(0,0)-$[x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…$

1.	The expansion of f(x,y), is
A.	f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+….\)
B.	f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]+…\)
C.	f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…\)
D.	f(0,0)-\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}]-…\)
Answer» C. f(0,0)+\([x \frac{∂f}{∂x}+y \frac{∂f}{∂y}]+\frac{1}{2!} [x^2 \frac{∂^2 f}{∂x^2}-2xy \frac{∂^2 f}{∂x∂y}-y^2 \frac{∂^2 f}{∂y^2}]+…\)

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