Why does ${\partial\over\partial t} w_i(tx)=\sum_{j=1}^n{\partial\over\partial x_j}w_i(tx)x_j$?

Question

Let $w$ be a $1$-differntial form. Why does the equality holds? $${\partial\over\partial t} w_i(tx)=\sum_{j=1}^n{\partial\over\partial x_j}w_i(tx)x_j$$

Answer 1

Let $z=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)=t\,x=(t\,x_1,\dots,t\,x_n)$. For any function $f\colon U\to\Bbb R$, $U\subset\Bbb R^n$, \begin{align} \frac{\partial }{\partial t}f(t\,x)&=\frac{\partial }{\partial t}f(y)\\ &=\frac{\partial f(y) }{\partial y_1}\,\frac{\partial y_1}{\partial t}+\dots+\frac{\partial f(y) }{\partial y_n}\,\frac{\partial y_n}{\partial t}\\ &=\frac{\partial f(y) }{\partial y_1}\,x_1+\dots+\frac{\partial f(y) }{\partial y_n}\,x_n. \end{align} Now, $\dfrac{\partial f(y) }{\partial y_j}$ is just the partial derivative of $f$ with respect to the $j$-th variable evaluated at $y=t\,x$, which is $\dfrac{\partial f(t\,x) }{\partial x_j}$.