Question

find the taylor series of x power y near the point (1,1) up to first degree terms​

Answer 1

Answer: The answer to the above given question is

f ( x , y ) = 1 + ( x - 1 ).

Step-by-step explanation:

Generally for function f ( x , y ) Taylor's series expansion will be

f ( x , y ) = f ( a , b ) + [ ( x - a ) $\frac{df}{dx}$ + ( y - b ) $\frac{df}{dy}$ ] + .....

Given f ( x , y ) = $x^{y}$

and ( a, b ) = ( 1 , 1 )

derivation formulas are given below
d $x^{n}$ = n $x^{n-1}$

d ( $a^{x}$ ) = $a^{x}$ log a
log ( 1 ) = 0

f ₓ ( x , y ) = y $x^{y-1}$

fy ( x , y ) = $x^{y}$ logx
So f ( 1, 1 ) = 1

fₓ ( 1, 1 ) = 1

fy ( 1 , 1 ) = 0

substitute these values in general formula

f ( x , y ) = $x^{y}$

= 1 + $\frac{1}{1!}$ [ ( x - 1 ) ( 1 ) + ( y - 1 ) ( 0 ) ]

= 1 + ( x - 1 )

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Answer 2

Answer:

The answer to the above given question is f ( x , y ) = 1 + ( x - 1 ).

Step-by-step explanation:

Step 1: Generally for function f(x, y) Taylor's series expansion will be

$f(x, y)=f(a, b)+\left[(x-a)^{\frac{d f}{d x}}+(y-b)^{\frac{d f}{d y}}\right]+\ldots \ldots$

Given $\mathrm{f}(\mathrm{x}, \mathrm{y})=x^y$

and (a, b)=(1,1)

Step 2:Derivation formulas are given below

$\mathrm{d} x^n=\mathrm{n} x^{n-1}$

$d\left(a^x\right)=a^x \log a$

$\log (1)=0$

$\mathrm{f}_{\mathrm{x}}(\mathrm{x}, \mathrm{y})=\mathrm{y} x^{y-1}$

fy$(\mathrm{x}, \mathrm{y})=x^y \log x$

So f(1,1)=1

f_x(1,1)=1

fy (1,1)=0

Step 3: Substitute these values in general formula

$\begin{aligned}& f(x, y)=x^y \\& =1+\frac{1}{1 !}[(x-1)(1)+(y-1)(0)]\end{aligned}$

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