# Chain rule question justification

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 $$0$$.

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)