Derivative of vector-valued function's norm-squared

Asked May 10 '22 at 23:59

Active May 11 '22 at 00:42

Viewed 95 times

I'm trying to understand the partial chain rule and want to check my understanding.

Let $$g(\mathbf{x})=\Vert{\mathbf{f}(\mathbf{x})}\Vert^2=\mathbf{f}(\mathbf{x})^T\mathbf{f}(\mathbf{x})$$

Then am I right in saying $$\frac{\text{d}g}{\text{d}\mathbf{x}}=2(\mathbf{J}(\mathbf{f})(\mathbf{x}))^T\mathbf{f}(\mathbf{x})$$

edited May 11 '22 at 00:42

asked May 10 '22 at 23:59

user18898152

1

I have never seen that definition of $|f(x)|$. – Kurt G. May 11 '22 at 00:39
My apologies, I was missing a power of 2 – user18898152 May 11 '22 at 00:42
I believe you are right. – blamocur May 11 '22 at 00:51
Indeed, you can simply use the chain rule along with this answer to get this result – Lorago May 11 '22 at 00:51

0 Answers0