Multivariate calculus problem: Calculating the gradient vector

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SUMMARY

The discussion focuses on calculating the gradient vector for a function using multivariate calculus. The partial derivatives are determined as ##f_x = \frac{2\ln{(x)}}{x}## and ##f_y = \frac{2\ln{(y)}}{y}##, leading to the gradient vector ##\nabla{f} = \begin{bmatrix} \frac{2\ln{(x)}}{x} \\ \frac{2\ln{(y)}}{y} \end{bmatrix}##. Additionally, the derivative of the vector function ##\mathbf{r}(t) = \left(\cos{(t)}e^{\sin{(t)}},-\sin{(t)}e^{\cos{(t)}}\right)## is discussed. The application of the chain rule is emphasized, particularly in the context of deriving composite functions.

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squenshl
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Homework Statement
Let ##U := \left\{(x,y)\in \mathbb{R}^2: xy\neq 0\right\}## and let ##f: U\mapsto \mathbb{R}## be defined by
$$f(x,y) := (\log_{e}{(|x|)})^2+(\log_{e}{(|y|)})^2.$$


1. Calculate ##\nabla f(x,y)## at each point of ##U##.

2. Let ##\mathbf{r}: (0,1)\mapsto \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).##

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.
Relevant Equations
None
1. We find the partial derivatives of ##f## with respect to ##x## and ##y## to get ##f_x = \frac{2\ln{(x)}}{x}## and ##f_y = \frac{2\ln{(y)}}{y}.## This makes the gradient vector
$$\nabla{f} = \begin{bmatrix}
f_x \\
f_y
\end{bmatrix} = \begin{bmatrix}
\frac{2\ln{(x)}}{x} \\
\frac{2\ln{(y)}}{y}
\end{bmatrix}.$$

2. We have
$$\mathbf{r}'(t) = \left(\cos{(t)}e^{\sin{(t)}},-\sin{(t)}e^{\cos{(t)}}\right).$$

After this I'm a little confused. Any help is appreciated.
 
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For (3), what does your version of the chain rule say?

If your multivariable calculus textbook is rigorous, it might also want you to show that ##U## is an open set (which is easily seen by a drawing).
 
The derivative of ##\mathbf{r}## at each point of ##(0,1)##?
 
squenshl said:
The derivative of ##\mathbf{r}## at each point of ##(0,1)##?
That confused me initially too. It would be clearer if it said "at each point in the open interval (0,1)."
 
Okay (1) and (2) are done.
So for (3), assuming ##t > 0##, ##f\circ \mathbf{r} = \ln{(e^{\sin{(t)}})}^2+\ln{(e^{\sin{(t)}})}^2 = \sin^2{(t)}+\cos^2{(t)} = 1## so the derivative is ##0##.
 
For #3, you need calculate the derivative using the chain rule if it can be applied.
 
Okay then I’m lost how do we then justify whether to use the chain rule?
 
I'm sure the conditions are stated in your textbook or were covered in lecture. Look them up.
 

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