Directional Derivative - Right Steps?

[kawsar]

1. Find the directional derivative of the function f(x,y) = e^{xy} at the point (-2,0) in the direction of the unit vector that makes an angle of \pi/3 with the positive x-axis



The vector to make into a unit vector is (\sqrt{3},1) and e^{xy} differentiated with respect to x and y is both e^{xy} (I hope I'm right here!)

Made the unit vector (\frac{1}{2}(\sqrt{3},1))

Use the dot product of (e^{xy},e^{xy}) and (\frac{1}{2}(\sqrt{3},1))

Then use the coordinates from the point (-2,0) and to get \sqrt{3}+1

Is this correct or have I missed a step or two?

Thanks!

 

[spamiam]

1. Find the directional derivative of the function f(x,y) = e^{xy} at the point (-2,0) in the direction of the unit vector that makes an angle of \pi/3 with the positive x-axis

The vector to make into a unit vector is (\sqrt{3},1) and e^{xy} differentiated with respect to x and y is both e^{xy} (I hope I'm right here!)

Your mistake seems to be in taking the partial derivatives (as perhaps you suspected). Don't forget the chain rule!

For example,
<br /> \frac{\partial}{\partial x} \sin(x^2y) = 2xy \cos(x^2y)<br />

Other than that, everything else looks good.

 

[kawsar]

Hmmm... As I suspected!

Could you tell me the partial derivatives of e^xy please?

Thanks.

 

[SammyS]

Do you know the derivative of e^2x?

 

[kawsar]

2e^(2x) so the other one is y*e^(xy) and x*e^(xy)...

Am I correct?

 

[spamiam]

2e^(2x) so the other one is y*e^(xy) and x*e^(xy)...

Am I correct?

Looks good. So what is your new answer?

 

[SammyS]

2e^(2x) so the other one is y*e^(xy) and x*e^(xy)...

Am I correct?

You are if you mean:

\frac{\partial }{\partial x}(e^{xy})=ye^{xy}

and

\frac{\partial }{\partial y}(e^{xy})=xe^{xy}

 

[kawsar]

Looks good. So what is your new answer?

I get -1

Correct?

 

[SammyS]

I get -1

Correct?

Yes.