Transpose of a partial derivative I would really appreciate if you can help with the following equations.
My prof puts the following equation, followed by the equation below.
$$g(\dot{\mathbf{q}})=\frac{1}{2} \dot{\mathbf{q}}^{\top} W \dot{\mathbf{q}}+\lambda^{\top}(\dot{\mathrm{x}}-J \dot{\mathbf{q}})$$
$$\left(\frac{\partial g}{\partial \dot{\mathbf{q}}}\right)^{\top}=W \dot{\mathbf{q}}+J^{\top} \boldsymbol{\lambda}=\mathbf{0}$$
I know that we are taking the transpose of the partial derivative. Shouldn't the answer be $$\left(\frac{\partial g}{\partial \dot{\mathbf{q}}}\right)^{\top} = (W\dot{\mathbf{q}} - \lambda^TJ)^T.$$
I would really appreciate if anyone can guide me on this.
Thanks in advance.
 A: Notation $\frac{\partial f(\mathbf x)}{\partial \mathbf x}$ stands for a vector with components $\left(\frac{\partial f(\mathbf x)}{\partial x_1}, \dots, \frac{\partial f(\mathbf x)}{\partial x_n}\right)$. It is convenient to treat it like a row vector (think of it like a Jacobian matrix of 1-component vector function).
Let's see how $\frac{\partial}{\partial \mathbf x}$ acts on simple functions.
$$
\frac{\partial}{\partial \mathbf x} \left(\mathbf a^\top \mathbf x\right) = 
\left(
\frac{\sum_i a_i x_i}{\partial x_1},
\dots,
\frac{\sum_i a_i x_i}{\partial x_n}
\right) = 
(a_1, \dots, a_n) = \mathbf a^\top.
$$
Here $\mathbf a$ is some constant column vector. The answer is $\mathbf a^\top$ because the answer should be a row vector.
Note that $\mathbf a^\top \mathbf x$ is the same as $\mathbf x^\top \mathbf a$ (assuming that we're dealing with real-valued vectors). Thus
$$
\frac{\partial}{\partial \mathbf x} \left(\mathbf x^\top \mathbf a\right) = \mathbf a^\top.
$$
The same holds when $\mathbf x$ is multiplied with matrix:
$$\frac{\partial}{\partial \mathbf x} \left(A \mathbf x\right) = A$$
$$\frac{\partial}{\partial \mathbf x} \left(\mathbf x^\top B\right) = B^\top$$
The rule of thumb is
$$
\frac{\partial}{\partial \mathbf x} \left(\text{something} \cdot \mathbf x\right) = \text{something}\\
\frac{\partial}{\partial \mathbf x} \left(\mathbf x^\top \cdot \text{something}\right) = (\text{something})^\top\\
$$
Here is assumed that "$\text{something}$" does not depend on $\mathbf x$.
Let's now differentiate your function
$$
g(\dot{\mathbf q}) = \frac{1}{2} \dot{\mathbf q}^\top W \dot{\mathbf q} + \boldsymbol \lambda^\top (\dot{\mathbf x} - J\dot{\mathbf q}) = 
\frac{1}{2} \dot{\mathbf q}^\top W \dot{\mathbf q} -
\boldsymbol \lambda^\top J\dot{\mathbf q} +
\boldsymbol \lambda^\top \dot{\mathbf x}
$$
The quadratic term
$$
\frac{\partial}{\partial \dot{\mathbf q}}\left(
\dot{\mathbf q}^\top W \dot{\mathbf q}
\right) = (\text{product rule}) = \frac{\partial}{\partial \dot{\mathbf q}}\left(
\dot{\mathbf q}^\top \underbrace{W \dot{\mathbf q}}_\text{treat as constant term}
\right) + \frac{\partial}{\partial \dot{\mathbf q}}\left(
\underbrace{\dot{\mathbf q}^\top W}_\text{treat as constant term} \dot{\mathbf q}
\right) = \\ = (W \dot{\mathbf q})^\top + \dot{\mathbf q}^\top W = 
\dot{\mathbf q}^\top (W^\top + W).
$$
I assume that $W$ is a symmetric matrix, so the last derivative simplifies to $2\dot{\mathbf q}^\top W$.
The next term
$$
\frac{\partial}{\partial \dot{\mathbf q}}\left(
\boldsymbol \lambda^\top J\dot{\mathbf q}
\right) 
=
\boldsymbol \lambda^\top J.
$$
The last one does not depend on $\dot{\mathbf q}$ thus is zero.
Collecting it altogether gives
$$
\frac{\partial g(\dot{\mathbf q})}{\partial \dot{\mathbf q}} = 
\dot{\mathbf q}^\top W - \boldsymbol \lambda^\top J.
$$
Transposing both sides gives the answer
$$
\left(\frac{\partial g(\dot{\mathbf q})}{\partial \dot{\mathbf q}}\right)^\top = 
W \dot{\mathbf q} - J^\top \boldsymbol \lambda.
$$
Note the sign in front of $J$, your professor has the opposite.
