Problem on directional derivatives - Leading Lesson

Problem on directional derivatives

At a particular point, the differentiable function $f(x,y)$ has directional derivatives $$D_\mathbf{\bfu_1}f = 1 \text{, and } D_\mathbf{\bfu_2}f = -2$$ where $$\mathbf{\bfu_1} = \frac{1}{\sqrt{2}} \mathbf{i} - \frac{1}{\sqrt{2}} \mathbf{j} \text{, and } \mathbf{\bfu_2} = \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j}.$$

Find the value of $\partial_x f$ and $\partial_y f$.

Solution
We are trying to find the partial derivatives of a two-variable function given information on the directional derivatives.
To do that, we try to relate the partial derivatives of $f$ to the directional derivatives of $f$.
Recall that
Directional derivative and gradient

The directional derivative of $f$ in the direction of $\bfu$ can be computed by: $$
D_{\bfu} f = \nabla f \cdot \frac{\bfu}{|\bfu|}$$
Definition of the gradient

The gradient of $f$ is defined as: $$\nabla f = \partial_x f \ \bfi + \partial_y f \ \bfj$$
Length of a vector

We now write out equations given by the directional derivatives for $\bfu_1$ and $\bfu_2$. Noting that both $|\bfu_1|=1$ and $|\bfu_2| = 1$, we get
\begin{align}
1 &= (\partial_x f, \partial_y f)\cdot \left(\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right),\\
-2 &=(\partial_x f, \partial_y f)\cdot \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right).
\end{align}
Direct computation of dot product

Expanding the dot product and multiplying by $\sqrt{2}$,
\begin{align}
\sqrt{2} &= \partial_x f - \partial_y f, \\
-2 \sqrt{2} &= \partial_x f + \partial_y f.
\end{align}
Adding and subtracting these equations gives $$ \partial_x f = - \frac{1}{\sqrt{2}}, \qquad \partial_y f = -\frac{3}{\sqrt{2}}.$$

Related topics

Multivariable calculus
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