To find the derivative of the given cost function with respect to the parameters θ₀ and θ₁, we'll follow these steps. The cost function is:
\[ \text{Cost} = \frac{1}{2m} \sum_{i=1}^{m} (\theta_0 + \theta_1 x^{(i)} - y^{(i)})^2 \]This is the mean squared error cost function commonly used in linear regression, where:
Let's rewrite the cost function for clarity:
\[ \text{Cost} = \frac{1}{2m} \sum_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)})^2 \]where h_θ(x⁽ⁱ⁾) = θ₀ + θ₁x⁽ⁱ⁾ is the hypothesis function.
We want to find:
\[ \frac{\partial \text{Cost}}{\partial \theta_0} \]Using the chain rule:
\[ \frac{\partial \text{Cost}}{\partial \theta_0} = \frac{1}{2m} \sum_{i=1}^{m} 2(h_\theta(x^{(i)}) - y^{(i)}) \cdot \frac{\partial}{\partial \theta_0} (h_\theta(x^{(i)}) - y^{(i)}) \]Since ∂/∂θ₀(h_θ(x⁽ⁱ⁾) - y⁽ⁱ⁾) = 1, this simplifies to:
\[ \frac{\partial \text{Cost}}{\partial \theta_0} = \frac{1}{m} \sum_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \]We want to find:
\[ \frac{\partial \text{Cost}}{\partial \theta_1} \]Again, using the chain rule:
\[ \frac{\partial \text{Cost}}{\partial \theta_1} = \frac{1}{2m} \sum_{i=1}^{m} 2(h_\theta(x^{(i)}) - y^{(i)}) \cdot \frac{\partial}{\partial \theta_1} (h_\theta(x^{(i)}) - y^{(i)}) \]Since ∂/∂θ₁(h_θ(x⁽ⁱ⁾) - y⁽ⁱ⁾) = x⁽ⁱ⁾, this simplifies to:
\[ \frac{\partial \text{Cost}}{\partial \theta_1} = \frac{1}{m} \sum_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x^{(i)} \]The partial derivatives of the cost function with respect to θ₀ and θ₁ are:
\[ \frac{\partial \text{Cost}}{\partial \theta_0} = \frac{1}{m} \sum_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \]\[ \frac{\partial \text{Cost}}{\partial \theta_1} = \frac{1}{m} \sum_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x^{(i)} \]These derivatives are used in gradient descent to update the parameters θ₀ and θ₁ iteratively.