Question

Show that f(x, y) = 5x eˣʸ is differentiable at (1, 0) and find its linearization there. Then use it to approximate f(1.1, −0.1).

Answer 1

Answer: The approximate f(1.1,-0.1) is 4.92.

Explanation:

Since we have given that

`f(x,y)=5xe^(xy)`

at (1,0),

we get
`f(1,0)=5`

Partial derivative would be

`f_x=5e^(xy)+5xye^(xy)=5e^(xy)(1+xy)`

So,

`f_x(1,0)=5`

Similarly,

`f_y=5x^2ye^(xy)`

So,
`f_y(1,0)=5`

Now,

`L(x,y)=f(1,0)+f_x(1,0)(x-1)+f_y(1,0)(y-0)\\\\L(x,y)=5+5(x-1)+5y\\\\L(x,y)=5+5x-5+5y\\\\L(x,y)=5x+5y`

L(1.1,-0.1)=
`5(1.1)-5(0.1)=5`

and

f(1.1,-0.1) =
`5(1.1)e^(1.1* -0.1)=4.92`

Hence, the approximate f(1.1,-0.1) is 4.92.