Compute the gradient and the directional derivative in the direction of the vector (0, -1) for the function f(x, y) = x^2 y + x e^(-xy).

Step-by-step solution

1. 1

   Recall the rule and check the direction vector. For a unit vector $\mathbf v$, the directional derivative is the dot product $D_{\mathbf v}f=\nabla f\cdot \mathbf v$. Here $\|(0,-1)\|=\sqrt{0^2+(-1)^2}=1$, so $\mathbf v$ is already a unit vector and no normalisation is needed. Had it not been, we would have had to divide by its length first.

2. 2

   Differentiate with respect to $x$. The term $x^2y$ gives $2xy$. The term $xe^{-xy}$ needs the product rule together with the chain rule, since $\partial_x e^{-xy}=-y e^{-xy}$:

   $\frac{\partial}{\partial x}\left(xe^{-xy}\right)=e^{-xy}-xy\,e^{-xy}.$

   So

   $f_x=2xy+(1-xy)e^{-xy}.$

3. 3

   Differentiate with respect to $y$. Now $x$ is a constant, so $x^2y$ gives $x^2$, and $\partial_y e^{-xy}=-x e^{-xy}$ makes the second term $-x^2 e^{-xy}$:

   $f_y=x^{2}-x^{2}e^{-xy}=x^{2}\left(1-e^{-xy}\right).$

4. 4

   Assemble the gradient.

   $\nabla f(x,y)=\begin{pmatrix}2xy+(1-xy)e^{-xy}\\[2pt] x^{2}\left(1-e^{-xy}\right)\end{pmatrix}.$

5. 5

   Dot with the direction. Because $\mathbf v=(0,-1)$ kills the first component, only the $y$-partial survives, with a sign flip:

   $D_{\mathbf v}f=\nabla f\cdot(0,-1)=-f_y=-x^{2}\left(1-e^{-xy}\right)=x^{2}\left(e^{-xy}-1\right).$

   That is exactly what the geometry says: moving in the $-y$ direction is moving backwards along $y$, so the rate of change is the negative of $f_y$.

6. 6

   Spot-check numerically. At $(x,y)=(1.3,0.7)$ a central difference gives $f_x\approx 1.856227$ and $f_y\approx 1.009734$, matching the formulas, and $D_{\mathbf v}f\approx-1.009734$. At $(-2,1.5)$ the formula gives $D_{\mathbf v}f\approx 76.342$, again matching the numerical derivative.