1
$\begingroup$

Consider ${\bf x}$ and ${\bf y}$ to be column vectors, and $a=f({\bf x})$ to be a scalar function of ${\bf x}$. (Also ${\bf x}$ and ${\bf y}$ are mutually independent.)

I need to find $\frac{\partial {\bf y}^T a}{\partial {\bf x}}$.

If I consider this to be vector-vector calculus, I don't know how to proceed since https://en.wikipedia.org/wiki/Matrix_calculus#Scalar-by-vector_identities does not list vector identities where numerator is a row vector and denominator is a column vector.

If I proceed by pulling ${\bf y}^T$ out of the integral since it does depend on ${\bf x}$ (analogous to a constant), then $\frac{\partial a}{\partial {\bf x}}$ is a row vector and matrix multiplication by 2 row vectors is not allowed ...

Someone help me !!

$\endgroup$
1
$\begingroup$

Of course nobody can differentiate with respect to a vector variable, but some people denote the gradient with respect to the vector variable ${\bf x}$ by ${\partial\over\partial{\bf x}}.\ $ As $a:=f({\bf x})$ is a scalar function of ${\bf x}$ it makes sense to consider this gradient. One might argue that a gradient by its nature is a row vector, but anyway, it seems that the authors consider $\nabla f({\bf x})$ again as a column vector. Therefore it makes sense to consider the scalar $${\bf y}^\top\>\nabla f({\bf x})=\sum_{k=1}^n y_k\>{\partial f\over\partial x_k}({\bf x})\ .$$ Note that differentiation with respect to the $x_k$ does not affect the components of ${\bf y}$, which has to be considered as a constant vector here.

$\endgroup$

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.