What is fx(0,1) in Partial Differentiation for f(x,y)=2e^(x^2y)?

[Mitchtwitchita]

f(x,y)=2e^(x^2y), then fx(0,1) = ?

fx = 4xye^(x^2y)
=4(0)(1)e^(0)(1)
=e^0
=1?

I'm told that I'm getting this question wrong but don't know how. Can anybody please help show me what I'm doing wrong on this one?

 

[Gib Z]

You are doing partial differentiation with respect to x, so you can treat y as a constant.

It's just simply using the chain rule and $(e^x)' = e^x$.

If you do that correctly you should get up to $2 \cdot ( 2y \cdot x^{2y-1} ) \exp (x^{2y})$.

The x term is a nice fat zero, giving the final result of 0.

 

[Mitchtwitchita]

Crap! I see now. Thanks Gib.

 

[HallsofIvy]

f(x,y)=2e^(x^2y), then fx(0,1) = ?

fx = 4xye^(x^2y)
=4(0)(1)e^(0)(1)
=e^0
=1?

I'm told that I'm getting this question wrong but don't know how. Can anybody please help show me what I'm doing wrong on this one?

Is that
$$2e^{x^{2y}}$$
as Gib Z assumed or
$$2e^{x^2y}$$
which is what I would assume?
If it is the latter then
$$f_x= 4xye^{x^2y}$$
as you have. Then $f_x(0,1)= 4(0)(1)e^{(0)(1)}= 0$, not 1, because of the "0" multiplying the exponential.

 

[Gib Z]

Damn i just realized that :( Looking at his working it looks like you're right Halls =] Sorry guys.

 

[Mitchtwitchita]

It was the latter. But I realized what I was doing wrong from Gib's reply. Thanks guys!