0
$\begingroup$

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 $0$.

$\endgroup$
0
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ Thank you so much!!! $\endgroup$ – squenshl Jul 29 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.