Please check my partial derivative of function

Question

Please check my solution + help me with the partial derivation with respect to $\beta_2$

$$Q = \sum(y_i-\beta_1-\beta_2^x)^2$$

$$\frac{\partial Q}{\partial \beta_1} = 2\sum(y_i-\beta_1-\beta_2^x)\cdot(-1) = 2\left(-\sum y_i + \beta_1 + \beta_2^x\right) \text{ ?}$$

$$\frac{\partial Q}{\partial \beta_2} = 2\sum(y_i-\beta_1-\beta_2^x)\cdot(\text{?})$$

Thanks

Answer 1

I suppose that what you want to minimize is $$Q = \sum(y_i-\beta_1-\beta_2^{x_i})^2$$ So, $$\frac{dQ}{d\beta_1}=2\sum(y_i-\beta_1-\beta_2^{x_i})\times \frac d{d\beta_1}(y_i-\beta_1-\beta_2^{x_i})=-2\sum(y_i-\beta_1-\beta_2^{x_i})\tag 1$$ $$\frac{dQ}{d\beta_2}=2\sum(y_i-\beta_1-\beta_2^{x_i})\times \frac d{d\beta_2}(y_i-\beta_1-\beta_2^{x_i})=-2\sum(y_i-\beta_1-\beta_2^{x_i})x_i \beta_2^{x_i-1}\tag 2$$ From $(1)$ you could eliminate $\beta_1$ and then $(2)$ reduces to a nonlinear equation in $\beta_2$; you will need a numerical method to solve it.