Question

expand f(x,y)=e^xy in Taylor series at (1,1) upto second degree​

Answer 1

fₓ(x,y) = $e$ +  $e$(x-1) +  $e$(y-1) +  $e$/2  $(x-1)^{2}$ + $e$(x-1)(y-1) + $e$/2$(y-1)^{2}$

Given:

The function is f(x,y) = $e^{xy}$

We have to find expansion of f(x,y) = $e^{xy}$ about (1,1) of order 2.

Now finding the partial derivatives of the function at (1,1):

fₓ(x,y) = $ye^{xy}$

fₓ(1,1) = $e$

fₓ(x,y) = $xe^{xy}$

fₓ(1,1) = $e$

fₓ(x,y) = $y^{2} e^{xy}$

fₓ(1,1) = $e$

fₓ(x,y) = $y^{2} e^{xy}$ + $e^{xy}$

fₓ(1,1) = $e$ + $e$

         = 2$e$

fₓ(x,y) = $x^{2} e^{xy}$

fₓ(1,1) = $e$

Hence, the expansion of the function f(x,y) = $e^{xy}$ about (1,1) is given as,

fₓ(x,y) = f(1,1) + fₓ(1,1)(x-1) + fy(1,1)(y-1) + fₓx/2$(x-1)^{2}$ + fxy(1,1)(x-1)(y-1)+ fyy/2$(y-1)^{2}$

fₓ(x,y) = $e$ +  $e$(x-1) +  $e$(y-1) +  $e$/2  $(x-1)^{2}$ + $e$(x-1)(y-1) + $e$/2$(y-1)^{2}$

Hence, the expansion upto second degree is given as,

fₓ(x,y) = $e$ +  $e$(x-1) +  $e$(y-1) +  $e$/2  $(x-1)^{2}$ + $e$(x-1)(y-1) + $e$/2$(y-1)^{2}$

Answer 2

Answer:

Step-by-step explanation: