I've tryied to calculate the following derivative using the chain rule, but in vain. Could you please help me with it? So I have a absolute value of a gradient of a function f, and I need to calculate the derivative of it with respect to the same function f.
$$\frac{d(|\nabla f|)}{df}$$
thanks!
To correct my question.
I came to the $$\frac{d(|\nabla f|)}{df}$$ derivative mistakenly.
The expression that wanted to evalualte was $$\nabla e^{f(x)|\nabla f(x)|}$$, and thanks to @mfl now I know how to do it.
Partial answer
We have that
\begin{align} \dfrac{\partial}{\partial x_i}e^{f(x)|\nabla f(x)|}&=e^{f(x)|\nabla f(x)|}\dfrac{\partial}{\partial x_i}(f(x)|\nabla f(x)|) \\&=e^{f(x)|\nabla f(x)|}\left(|\nabla f(x)|\dfrac{\partial}{\partial x_i}f(x)+f(x)\dfrac{\partial}{\partial x_i}|\nabla f(x)|\right) \end{align}
Now, it is
$$|\nabla f|=\sqrt{\sum_{i=j}^n\left(\dfrac{\partial f}{\partial x_j}\right)^2}$$
and thus
$$\dfrac{\partial}{\partial x_i}|\nabla f(x)|=\dfrac{\displaystyle\dfrac{\partial}{\partial x_i}\sum_{j=1}^n\left(\dfrac{\partial f}{\partial x_j}\right)^2}{\displaystyle2\sqrt{\sum_{j=1}^n\left(\dfrac{\partial f}{\partial x_j}\right)^2}}=\dfrac{\displaystyle\sum_{j=1}^n\dfrac{\partial^2 f}{\partial x_i\partial x_j}}{\displaystyle\sqrt{\sum_{j=1}^n\left(\dfrac{\partial f}{\partial x_j}\right)^2}}.$$
And now I hope you can get the answer.