# Second partial derivatives (one seems strange)

I am trying to find the second-partial derivatives for the following equation:

$$g = \sum_{i=1}^n \left(y_i - \frac{\theta_1 x_i}{x_i+\theta_2}\right)^2$$

Here, $\theta_1$ and $\theta_2$ are the model parameters.

I start by finding the first-order partial derivatives:

1) $\frac{\partial g}{\partial\theta_1} = 2(y_i-\frac{\theta_1x_i}{x_i+ \theta_2})(\frac{-x_i}{x_i+\theta_2})$

2) $\frac{\partial g}{\partial\theta_2} = 2(y_i-\frac{\theta_1x_i}{x_i+ \theta_2})(\frac{-\theta_1x_i}{(x_i+\theta_2)^2})$

Then, I find the second-order partial derivatives:

1) $\frac{\partial g}{\partial\theta_1^2} = 2[(y_i-\frac{\theta_1x_i}{x_i+ \theta_2})(0)+(\frac{-x_i}{x_i+\theta_2})(\frac{-x_i}{x_i+\theta_2})] = 2(\frac{x_i}{x_i+\theta_2})^2$

2) $\frac{\partial g}{\partial\theta_2^2} = 2[(y_i-\frac{\theta_1x_i}{x_i+ \theta_2})(\frac{2\theta_1x_i}{(x_i+\theta_2)^3})+(\frac{\theta_1x_i}{(x_i+\theta_2)^2})(\frac{-\theta_1x_i}{(x_i+\theta_2)^2})] = 2[(y_i-\frac{\theta_1x_i}{x_i+ \theta_2})(\frac{2\theta_1x_i}{(x_i+\theta_2)^3})-(\frac{\theta_1x_i}{(x_i+\theta_2)^2})^2]$

3) $\frac{\partial g}{\partial\theta_1\theta_2} = \frac{-4}{(x_i+\theta_2)^3}$

I am a bit hesitant about my work, especially the way that second partial derivative for $\theta_2$ seems so much messy than that for $\theta_1$. If there is anything wrong with my work, I would really like to figure out how to correctly solve this!

I re-write the same function with different notations to write it easier as $$g=\left(y-\frac{p x}{q+x}\right)^2$$ where $p,q$ are variables. $$\frac{\partial g}{\partial p}=-\frac{2 x \left(y-\frac{p x}{q+x}\right)}{q+x}$$ $$\frac{\partial^2 g}{\partial p^2}=\frac{2 x^2}{(q+x)^2}$$ $$\frac{\partial g}{\partial q}=\frac{2 p x \left(y-\frac{p x}{q+x}\right)}{(q+x)^2}$$ $$\frac{\partial^2 g}{\partial q^2}=\frac{2 p^2 x^2}{(q+x)^4}-\frac{4 p x \left(y-\frac{p x}{q+x}\right)}{(q+x)^3}$$ $$\frac{\partial^2 g}{\partial p \partial q}=\frac{\partial^2 g}{\partial q\partial p}=\frac{2 x \left(y-\frac{p x}{q+x}\right)}{(q+x)^2}-\frac{2 p x^2}{(q+x)^3}$$
Note: your $\frac{\partial g}{\partial \theta_2}$ is incorrect, there is no "-" sign in the second bracket since $\theta_2$ is in the denominator.