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,
$$\frac{\partial}{\partial x} \sin(x^2y) = 2xy \cos(x^2y)$$

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.