I want to take first partial derivative w.r.t. $x_i$ (for $i,j,k=1,\ldots,n$) of $$x\mapsto\frac 1{|x-y|^{n-2}},\quad x\neq y.$$ where $y\in\mathbb{R}^n$ is fixed.

Can I ask here if the following calculation is correct? I would like to make sure because I need to take the second partial derivative w.r.t. $x_i$ as well. \begin{align} \frac{\partial}{\partial x_i} \left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_i} |x-y|^{2-n} \\ &= \frac{\partial}{\partial x_i} \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{\frac{2-n}2} \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} \frac d{dx_i} (x_i-y_i)^2 \\ &= (2-n) \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} (x_i-y_i) \\ &= (2-n) \frac{x_i-y_i}{|x-y|^n} \end{align}

Edit: What about the second partial derivative? (Edit: Yes, Jack is right, the $i$'s below should have been $k$'s.) \begin{align} \frac{\partial^2}{\partial^2 x_k} \left(\frac 1{|x-y|^{n-2}} \right) &= (2-n) \frac{\partial}{\partial x_k} \left(\frac{x_i-y_i}{|x-y|^n} \right) \\ &= (2-n) \left(\frac{|x-y|^n \frac{\partial}{\partial x_k}(x_i-y_i)-(x_i-y_i) \frac{\partial}{\partial x_k}|x-y|^{2n}}{|x-y|^{2n}} \right) \\ &= (2-n) \left(\frac{|x-y|^n (1)-(x_i-y_i) 2n|x-y|^{n-1}(x_k-y_k)}{|x-y|^{2n}} \right) \\ &= (2-n)\left(\frac{1}{|x-y|^n}-2n \frac{(x_i-y_i)(x_k-y_k)}{|x-y|^{n+1}} \right) \end{align}

Second edit: Following Jack's updated answer, I now have \begin{align} \frac{\partial^2}{\partial x_j \partial x_k}\left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_j} \left(\frac{x_k-y_k}{|x-y|^n} \right)\\ &= \frac{\partial}{\partial x_j}(x_k-y_k) \frac 1{|x-y|^n} + (x_k-y_k) \frac{\partial}{\partial x_j} \frac 1{|x-y|^n} \\ &= \frac 1{|x-y|^n} + n \frac{(x_j-y_j)(x_k-y_k)}{|x-y|^{n+2}}, \end{align} the last step owing to $(*)$ in Jack's answer with $n \mapsto n+2$.

  • $\begingroup$ The notation $i$ in the edited question is incorrect. $\endgroup$ – Jack May 25 '16 at 20:51
  • $\begingroup$ You wrote $j=k$ in $\partial^2/(\partial_j\partial x_k)$ before your editing. The answer would be different when $j\neq k$. $\endgroup$ – Jack May 25 '16 at 21:04
  • $\begingroup$ Give me a minute to edit my answer. $\endgroup$ – Jack May 25 '16 at 21:06
  • $\begingroup$ You calculation of $\partial/\partial x_j(x_k-y_k)$ is incorrect. $\endgroup$ – Jack May 25 '16 at 21:18

You calculation is basically correct except that you need to be careful with the dummy index.

\begin{align} \tag{*} \frac{\partial}{\partial x_k} \left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_k} |x-y|^{2-n} \\ &= \frac{\partial}{\partial x_k} \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{\frac{2-n}2} \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2}\frac {\partial}{\partial x_k} \left(\sum_{i=1}^n (x_i-y_i)^2 \right) \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} \sum_{i=1}^n\frac{\partial}{\partial x_k} \left( (x_i-y_i)^2 \right) \\ &= \frac{2-n}2 \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} 2(x_k-y_k) \\ &= (2-n) \left(\sum_{i=1}^n (x_i-y_i)^2 \right)^{-\frac{n}2} (x_k-y_k) \\ &=(2-n)\frac{x_k-y_k}{|x-y|^{n}} \end{align}

Now to find $\frac{\partial^2}{\partial x_k^2} \left(\frac 1{|x-y|^{n-2}} \right) $ it suffices by linearity to find $$ \frac{\partial}{\partial x_k}\left(\frac{x_k-y_k}{|x-y|^{n}}\right). $$ The product rule tells you that the nontrivial part is $$ \frac{\partial}{\partial x_k}\left(\frac{1}{|x-y|^{n}}\right)\tag{**} $$ But you can use $(*)$ to get $(**)$.

Here is another way to simplify the calculation (and notations) to get $(*)$. First notice that $$ \frac{\partial}{\partial x_k}|x-y|^2=2(x_k-y_k). $$ Then \begin{align} \frac{\partial}{\partial x_k} \left(\frac 1{|x-y|^{n-2}} \right) &= \frac{\partial}{\partial x_k} (|x-y|^2)^{\frac{2-n}{2}}\\ &=\frac{2-n}{2} (|x-y|^2)^{\frac{-n}{2}}\cdot 2(x_k-y_k) \end{align} Similarly, you can find $(**)$.

[Added due to the change in OP] Now to find $\frac{\partial^2}{\partial x_m\partial x_k} \left(\frac 1{|x-y|^{n-2}} \right) $ it suffices by linearity to find $$ \frac{\partial}{\partial x_m}\left(\frac{x_k-y_k}{|x-y|^{n}}\right)= (x_k-y_k)\frac{\partial}{\partial x_m}\left(\frac{1}{|x-y|^{n}}\right)+ \delta_{mk}\left(\frac{1}{|x-y|^{n}}\right) $$ where $\delta_{mk}=1$ when $m=k$ and $\delta_{mk}=0$ when $m\neq k$. The nontrivial part is $$ \frac{\partial}{\partial x_m}\left(\frac{1}{|x-y|^{n}}\right) =\frac{\partial}{\partial x_m}(|x-y|^2)^{(-n/2)}\\ =\frac{-n}{2}(|x-y|^2)^{-\frac{n}{2}-1}2(x_m-y_m)\\ =-n(x_m-y_m)|x-y|^{-n-2} $$


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.