Let $f(x)$ be a defined function $\mathbb{R}^2$ using polar coordinate by $\frac{r^6}{(\log(r))^3}(1+\cos(\theta))$ when $r>1$.
Please help me to calculate $\partial_x^{\alpha}f(x)$ for all $x\in\mathbb{R}^2$ and all $\alpha\in\mathbb{N}^2$ with $1\le |\alpha|\le 3$.
Merci de m'aider
On
I don't think there's a non-tedious way to solve this problem, so I'll try to outline what may be the least tedious. Let's first note$$\frac{\partial}{\partial r}=\left(\frac{\partial x}{\partial r}\right)_\theta\partial_x+\left(\frac{\partial y}{\partial r}\right)_\theta\partial_y=\cos\theta\partial_x+\sin\theta\partial_y$$and similarly $\partial_\theta=r(-\sin\theta\partial_x+\cos\theta\partial_y)$, results we can rearrange to$$\partial_x=\cos\theta\partial_r-\frac{\sin\theta}{r}\partial_\theta,\,\partial_y=\sin\theta\partial_r+\frac{\cos\theta}{r}\partial_\theta.$$We can now express$$\partial_x\,\partial_y,\,\partial_x^2,\,\partial_x\partial_y,\,\partial_y^2,\,\partial_x^3,\,\partial_x^2\partial_y,\,\partial_x\partial_y^2,\,\partial_y^3$$in terms of $r,\,\theta,\,\partial_r,\,\partial_\theta$. This saves us rewriting $f$ in terms of Cartesian variables. This will at times require differentiating $f$ up to three times with respect to $r$ (or $\theta$, but that's trivial), viz.$$\frac{\partial_rf}{1+\cos\theta}=\frac{3r^5(2\ln r-1)}{\ln^4r},\\\,\frac{\partial_r^2f}{1+\cos\theta}=\frac{3r^4(10\ln^2r-11\ln r+4)}{\ln^5r},\\\,\frac{\partial_r^3f}{1+\cos\theta}=\frac{6r^3(20\ln^3r-37\ln^2r+30\ln r-10)}{\ln^6r}.$$
The function and the variable you are deriving with respect to, must be compatible i.e. either you must change $\partial _x$ to $\partial _r$ and $\partial _\theta$ or you must write the function in cartesian coordinates and it'll be $$f(x,y)= \frac{(x^2+y^2)^3}{(\log (\sqrt {x^2+y^2}))^3}(1+\cos (\arctan (\frac{y}{x})))$$
Or write $\partial _x= \frac{\partial r}{\partial x}\partial_r + \frac{\partial \theta}{\partial x}\partial_\theta$