# 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.

• The answer cannot be what you suggest because $W\mathbf q$ is a column vector and $\boldsymbol\lambda^\top J$ is a row vector. Commented Jul 18, 2020 at 18:51
• @uranix, thanks for replying. I am not following what you saying. Could you show me the steps in between the first equation and the correct answer? Commented Jul 18, 2020 at 19:18

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.

• Thank you so much @uranix. I totally forgot that W is symmetric in this case! Commented Jul 19, 2020 at 1:28
• Is this correct? $$(\dot{\mathbf q}^\top W - \boldsymbol \lambda^\top J)^T = W \dot{\mathbf q} - J^\top \boldsymbol \lambda.$$ Commented Jul 19, 2020 at 1:32