Let $U := \left\{(x,y)\in \mathbb{R}^2: xy\neq 0\right\}$ and let $f: U \to \mathbb{R}$ be defined by \begin{equation*} f(x,y) := (\log_{e}{(|x|)})^2+(\log_{e}{(|y|)})^2. \end{equation*} 1. Calculate $\nabla f(x,y)$ at each point of $U$.
We find the partial derivatives of $f$ with respect to $x$ and $y$ to get $\frac{\partial f}{\partial x} = f_x = \frac{2\ln{(x)}}{x}$ and $\frac{\partial f}{\partial y} = f_y = \frac{2\ln{(y)}}{y}$. This makes the gradient vector \begin{equation*} \nabla{f} = \begin{bmatrix} f_x \\ f_y \end{bmatrix} = \begin{bmatrix} \frac{2\ln{(x)}}{x} \\[6pt] \frac{2\ln{(y)}}{y} \end{bmatrix}. \end{equation*} 2. Let $\mathbf{r}: (0,1) \to \mathbb{R}^2$ be defined by $\mathbf{r}(t) := \left(e^{\sin{(t)}},e^{\cos{(t)}}\right)$.
Calculate the derivative of $\mathbf{r}$ at each point of $(0,1)$.
We have \begin{equation*} \mathbf{r}'(t) = \left(\cos{(t)}e^{\sin{(t)}},-\sin{(t)}e^{\cos{(t)}}\right). \end{equation*}
3. Justify whether you can use the chain rule to calculate the derivative of $f\circ \mathbf{r}$.
If it is justifiable, calculate the derivative of $f\circ \mathbf{r}$ using the chain rule.
I'm having trouble with this one. Some help would be great!!!
Do we just say that since $f$ and $\mathbf{r}$ are both differentiable, $f\circ \mathbf{r}$ must be as well so we can apply the chain rule???
I know I haven't explicitly used the chain rule here but we have, for $t > 0$, \begin{equation*} f\circ \mathbf{r} = \ln{(e^{\sin{(t)}})}^2+\ln{(e^{\sin{(t)}})}^2 = \sin^2{(t)}+\cos^2{(t)} = 1. \end{equation*} Hence, the derivative is $0$.


1 Answer 1


In your partial derivatives of $f$, you need absolute values inside the logarithm: \begin{align} \dfrac{\partial f}{\partial x}(x,y) = \dfrac{2 \ln(|x|)}{x} \quad \text{and} \quad \dfrac{\partial f}{\partial y}(x,y) = \dfrac{2 \ln(|y|)}{y}. \end{align} For your last part about differentiability of $f \circ \boldsymbol{r}$, what you wrote is corrct, but perhaps it would be good to be more specific and mention that for every $t \in (0,1)$, $\boldsymbol{r}(t)$ lies inside the domain $U$ of $f$. I say this because the actual hypothesis of the chain rule is that $\boldsymbol{r}$ must be differentiable at $t$ and $f$ needs to be differentiable at $\boldsymbol{r}(t)$.

By the way, to use the chain rule here, you just need to calculate \begin{align} (f \circ \boldsymbol{r})'(t) = \nabla f(\boldsymbol{r}(t)) \cdot \boldsymbol{r}'(t) \end{align} (the $\cdot$ being matrix multiplication)

  • $\begingroup$ Thank you so much!!! $\endgroup$
    – squenshl
    Commented Jul 29, 2019 at 22:01
  • $\begingroup$ It looks as though you have forgotten to accept the answer... :) $\endgroup$ Commented Oct 28, 2021 at 22:56

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