Can anyone give me an expression for following differential problem:
$$ \frac{\partial f_1(X)^Tf_2(X)f_3(X)}{\partial X} = ?$$
where $ X $ is a matrix, $ f_1(X) $ is a vector, $f_2(X)$ is a matrix, and $ f_3(X)$ is a vector, so $f_1(X)^Tf_2(X)f_3(X)$ is a scalar.
Thank you so much
This is just a repeated application of the product rule.
Let $\phi(x) = \phi_1(x) \phi_2(x) \phi_3(x)$, and $\eta(x) = \phi_1(x) \phi_2(x)$. Then $\phi(x) = \eta(x) \phi_3(x)$.
Then, using the product rule, we have $D\phi(x)(h) = D\eta(x)(h) \phi_3(x)+ \eta (x) D\phi_3(x)(h)$ and $D\eta(x)(h) = D\phi_1(x)(h) \phi_2(x)+ \phi_1(x) D\phi_2(x)(h)$.
Combining gives: $D\phi(x)(h) = ( D\phi_1(x)(h) \phi_2(x)+ \phi_1(x) D\phi_2(x)(h) ) \phi_3(x)+ \phi_1(x) \phi_2(x) D\phi_3(x)(h)$, and so $$D\phi(x)(h) = D\phi_1(x)(h) \phi_2(x) \phi_3(x)+ \phi_1(x) D\phi_2(x)(h) \phi_3(x)+ \phi_1(x) \phi_2(x) D\phi_3(x)(h)$$ If $\phi_1(x) = f_1(x)^T$, then since $X \mapsto x^T$ is a linear operation, we have $D\phi_1(x)(h) = (Df(x)(h))^T$.
Letting $f_i(x) = \phi_i(x)$ for $i=2,3$ and substituting gives: $$Df(x)(h) = (Df(x)(h))^T f_2(x) f_3(x)+ f_1(x)^T Df_2(x)(h) f_3(x)+ f_1(x)^T f_2(x) Df_3(x)(h)$$