# Derivative of the square root of a diagonal matrix

Let the entries of the diagonal matrix $A \in R^{n\times n}$ are differentiable functions of $\theta$.

The matrix $A$ is:

\begin{bmatrix} \theta & 0 & 0 \\ 0 & \theta^3 & 0 \\ 0 & 0 & sin(\theta) \end{bmatrix}

What is the derivative of $\sqrt{A}$ :

$$\frac{\partial \sqrt{A}}{\partial \theta} = ?$$

• Is $\theta>0$? ${}$ – copper.hat Jan 9 '17 at 17:30
• yes copper.hat $\theta > 0$. – B.Mohammed Jan 10 '17 at 10:29

We have $$A=\text{diag }(\theta, \theta^3, \sin \theta)\Rightarrow \sqrt{A}=\text{diag }\left(\theta^{1/2}, \theta^{3/2}, \left(\sin \theta\right)^{1/2}\right)$$ $$\Rightarrow \frac{d}{d\theta}\sqrt{A}=\text{diag }\left(\frac{d}{d\theta}\theta^{1/2}, \frac{d}{d\theta}\theta^{3/2}, \frac{d}{d\theta}\left(\sin \theta\right)^{1/2}\right)=\ldots$$ for all $\theta\in\mathbb{R}$ such that the three derivatives exist.