I'm confused by multiple representations of the partial derivatives of Linear Regression cost function.
This is the MSE cost function of Linear Regression. Here $h_\theta(x) = \theta_0+\theta_1x$ .
\begin{aligned}J(\theta_0,\theta_1) &= \frac{1}{m}\displaystyle\sum_{i=1}^m(h_\theta(x^{(i)}) - y^{(i)})^2\\J(\theta_0,\theta_1) &= \frac{1}{m}\displaystyle\sum_{i=1}^m(\theta_0 + \theta_1x^{(i)} - y^{(i)})^2\end{aligned}
Are these the correct partial derivatives of above MSE cost function of Linear Regression with respect to $\theta_1, \theta_0$? If there's any mistake please correct me.
\begin{aligned}\frac{dJ}{d\theta_1} &= \frac{-2}{m}\displaystyle\sum_{i=1}^m(x^{(i)}).(\theta_0 + \theta_1x^{(i)} - y^{(i)})\\ \frac{dJ}{d\theta_0} &= \frac{-2}{m}\displaystyle\sum_{i=1}^m(\theta_0 + \theta_1x^{(i)} - y^{(i)})\end{aligned}
The derivatives are almost correct, but instead of a minus sign, you should have a plus sign. The minus sign is there if we differentiate
$$J = \dfrac{1}{m}\sum_{i=1}^m\left[y_i-\theta_0-\theta_1 x_i\right]^2$$
If we calculate the partial derivatives we obtain
$$\dfrac{\partial J}{\partial \theta_0}=\frac{2}{m}\sum_{i=1}^{m}[y_i-\theta_0-\theta_1x_i]\cdot\left[-1 \right]$$ $$\dfrac{\partial J}{\partial \theta_1}=\frac{2}{m}\sum_{i=1}^{m}[y_i-\theta_0-\theta_1x_i]\cdot\left[-x_i \right]$$
In order to find the extremum of the cost function $J$ (we seek to minimize it) we need to set these partial derivatives equal to $0$ $$\dfrac{\partial J}{\partial \theta_0}=\frac{2}{m}\sum_{i=1}^{m}[y_i-\theta_0-\theta_1x_i]\cdot\left[-1 \right]=0$$ $$\implies \sum_{i=1}^{m}[y_i-\theta_0-\theta_1x_i]=0$$ $$\dfrac{\partial J}{\partial \theta_1}=\frac{2}{m}\sum_{i=1}^{m}[y_i-\theta_0-\theta_1x_i]\cdot\left[-x_i \right]=0$$ $$\implies \sum_{i=1}^{m}[y_i-\theta_0-\theta_1x_i]\cdot\left[x_i\right] = 0.$$
As we divide by $ -2/m $ for both cases we will obtain the same result. If you had $ +2/m $ then you would divide by $ 2/m $ and still obtain the same equations as stated above. If the equation that we need to solve are identical the solutions will also be identical.