[proofplan] We prove equality of one-forms by evaluating both sides at an arbitrary point $p \in W$ and on an arbitrary tangent vector $v \in T_pM$. The chart $x$ gives the coordinate basis $\left(\frac{\partial}{\partial x_i}\big|_p\right)_{i=1}^n$ for $T_pM$ and the dual coframe $(dx_i|_p)_{i=1}^n$ for $T_p^*M$. Writing $v$ in this basis reduces the identity to the coordinate formula for the differential of the transition component $y_a \circ x^{-1}: x(W) \to \mathbb{R}$. [/proofplan] [step:Fix a point and express a tangent vector in the $x$-coordinate basis] Let $p \in W$ and let $v \in T_pM$. For each $i \in \{1,\dots,n\}$, let \begin{align*} \frac{\partial}{\partial x_i}\bigg|_p \in T_pM \end{align*} denote the coordinate tangent vector determined by the chart $(U,x)$. Since these vectors form a basis of $T_pM$, there exist unique [real numbers](/page/Real%20Numbers) $v_1,\dots,v_n \in \mathbb{R}$ such that \begin{align*} v = \sum_{i=1}^n v_i \frac{\partial}{\partial x_i}\bigg|_p. \end{align*} The coordinate one-forms $dx_1|_p,\dots,dx_n|_p \in T_p^*M$ are dual to this basis, hence \begin{align*} dx_j|_p(v)=v_j \end{align*} for every $j \in \{1,\dots,n\}$. [guided] Fix $p \in W$ and $v \in T_pM$. The chart $(U,x)$ identifies tangent vectors at $p$ with directional derivatives in the coordinate variables $x_1,\dots,x_n$. Thus the coordinate tangent vectors \begin{align*} \frac{\partial}{\partial x_1}\bigg|_p,\dots, \frac{\partial}{\partial x_n}\bigg|_p \end{align*} form a basis of the [vector space](/page/Vector%20Space) $T_pM$. Therefore there are unique coefficients $v_1,\dots,v_n \in \mathbb{R}$ such that \begin{align*} v = \sum_{i=1}^n v_i \frac{\partial}{\partial x_i}\bigg|_p. \end{align*} The one-forms $dx_1|_p,\dots,dx_n|_p$ are the dual coordinate covectors. This means that for every pair $i,j \in \{1,\dots,n\}$, \begin{align*} dx_j|_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right)=\delta_{ij}, \end{align*} where $\delta_{ij}$ is the Kronecker delta. Applying $dx_j|_p$ to the expansion of $v$ gives \begin{align*} dx_j|_p(v) = dx_j|_p\left(\sum_{i=1}^n v_i \frac{\partial}{\partial x_i}\bigg|_p\right) = \sum_{i=1}^n v_i \delta_{ij} = v_j. \end{align*} Thus the coefficients of $v$ in the $x$-coordinate basis are exactly its evaluations under the coordinate one-forms. [/guided] [/step] [step:Evaluate $dy_a$ on the coordinate expansion of $v$] Fix $a \in \{1,\dots,n\}$. Define the coordinate representation of the function $y_a|_W: W \to \mathbb{R}$ in the chart $x$ by the smooth map \begin{align*} Y_a: x(W) &\to \mathbb{R} \\ r &\mapsto y_a(x^{-1}(r)). \end{align*} By the definition of coordinate partial derivatives, \begin{align*} dy_a|_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right) = \frac{\partial Y_a}{\partial r_i}(x(p)) = \frac{\partial y_a}{\partial x_i}(p). \end{align*} Using the linearity of the covector $dy_a|_p:T_pM\to\mathbb{R}$, we obtain \begin{align*} dy_a|_p(v) &= dy_a|_p\left(\sum_{i=1}^n v_i \frac{\partial}{\partial x_i}\bigg|_p\right) \\ &= \sum_{i=1}^n v_i\,dy_a|_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right) \\ &= \sum_{i=1}^n v_i\,\frac{\partial y_a}{\partial x_i}(p). \end{align*} [guided] Fix $a \in \{1,\dots,n\}$. To compute $dy_a|_p(v)$, we rewrite the coordinate function $y_a$ in the $x$-coordinates. Define \begin{align*} Y_a: x(W) &\to \mathbb{R} \\ r &\mapsto y_a(x^{-1}(r)). \end{align*} This is the $a$-th component of the transition map from $x$-coordinates to $y$-coordinates. By definition, the coordinate vector $\frac{\partial}{\partial x_i}\big|_p$ differentiates a smooth function in the $i$-th $x$-coordinate direction. Therefore \begin{align*} dy_a|_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right) = \frac{\partial Y_a}{\partial r_i}(x(p)). \end{align*} The function $\frac{\partial y_a}{\partial x_i}:W\to\mathbb{R}$ was defined precisely by \begin{align*} \frac{\partial y_a}{\partial x_i}(p) = \frac{\partial Y_a}{\partial r_i}(x(p)). \end{align*} Hence \begin{align*} dy_a|_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right) = \frac{\partial y_a}{\partial x_i}(p). \end{align*} Now use the expansion \begin{align*} v = \sum_{i=1}^n v_i \frac{\partial}{\partial x_i}\bigg|_p \end{align*} and the linearity of the covector $dy_a|_p:T_pM\to\mathbb{R}$. This gives \begin{align*} dy_a|_p(v) &= dy_a|_p\left(\sum_{i=1}^n v_i \frac{\partial}{\partial x_i}\bigg|_p\right) \\ &= \sum_{i=1}^n v_i\,dy_a|_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right) \\ &= \sum_{i=1}^n v_i\,\frac{\partial y_a}{\partial x_i}(p). \end{align*} This is the left-hand side of the desired identity evaluated on the arbitrary tangent vector $v$. [/guided] [/step] [step:Evaluate the proposed right-hand side on the same tangent vector] Define the covector $\omega_p \in T_p^*M$ by \begin{align*} \omega_p := \sum_{i=1}^n \frac{\partial y_a}{\partial x_i}(p)\, dx_i|_p. \end{align*} Using the linearity of each $dx_i|_p:T_pM\to\mathbb{R}$ and the identity $dx_i|_p(v)=v_i$, we compute \begin{align*} \omega_p(v) &= \left(\sum_{i=1}^n \frac{\partial y_a}{\partial x_i}(p)\, dx_i|_p\right)(v) \\ &= \sum_{i=1}^n \frac{\partial y_a}{\partial x_i}(p)\, dx_i|_p(v) \\ &= \sum_{i=1}^n \frac{\partial y_a}{\partial x_i}(p)\, v_i. \end{align*} This agrees with the value of $dy_a|_p(v)$ computed above. [/step] [step:Conclude equality of the one-forms on the overlap] For the fixed $p \in W$ and fixed $a \in \{1,\dots,n\}$, the two covectors \begin{align*} dy_a|_p \quad\text{and}\quad \sum_{i=1}^n \frac{\partial y_a}{\partial x_i}(p)\, dx_i|_p \end{align*} have the same value on every $v \in T_pM$. Therefore they are equal in $T_p^*M$. Since $p \in W$ was arbitrary, the equality holds pointwise on $W$: \begin{align*} dy_a = \sum_{i=1}^n \frac{\partial y_a}{\partial x_i}\, dx_i. \end{align*} Because $a \in \{1,\dots,n\}$ was arbitrary, the formula holds for every coordinate one-form $dy_a$ on $U\cap V$. [/step]