Linear Approximation Homework: f(x,y)=(xe^y)^8

[KingBigness]

Homework Statement

Let f(x,y) = $(xe^y)^8$

i) Find

$\frac{∂f}{∂x}$ $\frac{∂f}{∂y}$ $\frac{∂^2f}{∂x^2}$

ii) Using a tangent plane of f(x,y) find an approximate value of (0.98e^0.01)^8

Homework Equations

The Attempt at a Solution

i)
$\frac{∂f}{∂x}$ =$8e^{8y}x^{7}$

$\frac{∂f}{∂y}$ = $8x^{8}e^{8y}$

$\frac{∂^2f}{∂x^2}$ = $56e^{8y}x^{6}$


ii) I have done many questions on finding linear approximations but I have always had a function, a point to evaluate the function at and points to approximate it at.

In this I have the function Let f(x,y) = $(xe^y)^8$ and want to use it to approximate f(0.98,0.01) but I'm not sure at what point I should evaluate it at.

Can anyone help out?

 

[HallsofIvy]

Since .98 is reasonably close to 1 and 0.01 close to 0, I think x= 1, y= 0 would be a good try.

 

[KingBigness]

Was leaning towards that, just wanted to make sure.

That all worked out nicely, thanks for your advice =D