# Directional derivative $f(x,y) = x^3+y^2 -2xy$ and $a=(1,1)^T$

Let $$f:\mathbb{R^2} \to \mathbb{R}$$ with $$f(x,y) = x^3+y^2 -2xy$$ and $$a=(1,1)^T$$.

How can one find out the directional derivative of $$f$$ in $$a$$ in the direction of northwest with the definition and with the help of the gradient?

$$\text{grad}(f) = \left(\begin{array}{c}3x^2-2y \\2y-2x\end{array}\right)$$

I don't know if this is correct, but to get the directional derivative one has to put the values of the gradient in the point $$a=(1,1)^T$$, so we get

$$\text{grad}(f)(1,1) = \left(\begin{array}{c}1 \\0 \end{array}\right)$$

Transposing it gives

$$\left(\begin{array}{c}1 & 0 \end{array}\right)$$

I have read that northwest is $$r = \frac{1}{\sqrt{2}} \left(\begin{array}{c}-1\\1\end{array}\right)$$

Then we look if the length of the vector is $$1$$.

$$|r| = 2$$

$$2 \neq 1$$, so we normalize the vector via

$$v = \frac{1}{|r|} \cdot r = \frac{1}{2} \cdot 2 = 1$$

And then we calculate the dot product with

$$D_vf(a) = \text{grad}(f)(1,1) \cdot v$$

$$= \left(\begin{array}{c}1 \\0 \end{array}\right) \cdot 1$$

$$= 1$$

I think this is wrong, but I just follwed the "algorithm"..

• The gradient of $f$ only has 2 entries, since it is defined on $\Bbb R^2$, so $\nabla(f)=grad(f)={3x^2-2y \choose 2y-2x}$ Commented Jan 14, 2020 at 13:59
• @Norse Sorry, my bad. Edited. Commented Jan 14, 2020 at 14:01

Actually, the gradient of $$f$$ at a point $$(x,y)$$ is $$\left[\begin{smallmatrix}3x^2-2y\\2y-2x\end{smallmatrix}\right]$$. So, the gradient of $$f$$ at $$\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$$ is $$\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$$. Since northeast is (indeed) $$\frac1{\sqrt2}\left[\begin{smallmatrix}-1\\1\end{smallmatrix}\right]$$, the partial derivative in that direction is$$\left\langle\begin{bmatrix}1\\0\end{bmatrix},\frac1{\sqrt2}\begin{bmatrix}-1\\1\end{bmatrix}\right\rangle=-\frac1{\sqrt2}.$$
• No, it is not. ${}$ Commented Jan 14, 2020 at 14:22