# Matrix Calculus- transpose

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 !!

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.