Directional Derivatives using Polar Coordinates I am having a hard time with this problem on my homework assignment.  Here is the problem, and i will show my work below:
If $f( x, y) = -2 x^{2} + 3 y^{2}$, find the value of the directional derivative at the point $( -1, 1)$ in the direction given by the angle $\theta = \frac{2 \pi}{2}$. More specifically, find the directional derivative of f at the point $\left(-1,1\right)$ in the direction of the unit vector determined by the angle $\theta$ in polar coordinates.
To find the directional derivative, i know you need to find the partial derivative with respect to x, and then to y as shown below:
$$fx(x,y)=-4x$$ $$fy(x,y)=6y$$
Then because we want the directional derivative at point P, we evaluate at point P:
$$\nabla f=4i+6j$$
Now the next step is to multiply the gradient by the directional vector, and that will give the final directional derivative.  I just don't know what the hint is implying when is says use the polar coordinates to find the direction of the unit vector.  What i do know about polar coordinates is:
$$x=rcos \theta$$ $$y=rsin\theta$$
But after that i don't know how to use it.  Any help? Thanks!

The unit vector determined by the angle $\theta$ in polar coordinates is:
$$\cos\theta\; i + \sin\theta\; j = \cos\pi\; i + \sin\pi\; j = -1\;i + 0\;j$$
I think you must have the projection
of $\nabla f=4i+6j$ on this unit vector: $$(4i+6j)\cdot(-1i+0j)\times(-1i+0j) = 4i+0j$$ Here $(\cdot)$ is the inner product and $(\times)$ the scalar times vector product.
