diff --git a/logistic-regressor-scratch/main.ipynb b/logistic-regressor-scratch/main.ipynb
index d3c2dab..f79626d 100644
--- a/logistic-regressor-scratch/main.ipynb
+++ b/logistic-regressor-scratch/main.ipynb
@@ -268,7 +268,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 3,
+   "execution_count": null,
    "id": "710dded0",
    "metadata": {},
    "outputs": [],
@@ -277,7 +277,7 @@
     "y = df['target'].to_numpy().reshape(-1, 1)\n",
     "\n",
     "# Convert target to binary: 1 if setosa (class 0), 0 otherwise\n",
-    "y = (y==0).astype(float)\n"
+    "y = (y==0).astype(float)"
    ]
   },
   {
@@ -325,6 +325,18 @@
     "*(Note: `np.clip` is used to bound extreme values and prevent overflow errors during exponential calculation).*"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "c302e5fe",
+   "metadata": {},
+   "source": [
+    "### Sigmoid Function Base Behavior\n",
+    "\n",
+    "This plot isolates the **Sigmoid function's behavior** in isolation, independent of the model's training optimization. \n",
+    "\n",
+    "It serves exclusively as a mathematical baseline to demonstrate the activation range between $0$ and $1$ and the function's sensitivity to input feature values. This visualization does not represent the optimized decision boundary or the final classification results; rather, it highlights the function's intrinsic mapping capability before weight adjustments are applied."
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 5,
@@ -399,6 +411,47 @@
     "Probabilities are clipped using a tiny epsilon ($\\epsilon$) to prevent mathematical undefined errors (like $\\log(0)$), which would break the algorithm."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "ec6509c3",
+   "metadata": {},
+   "source": [
+    "### Derivation of the Log Loss (Binary Cross-Entropy) Cost Function\n",
+    "\n",
+    "In binary classification models, the goal is to estimate the probability that an instance belongs to the positive class. To optimize the model, we need a cost function that heavily penalizes confident but incorrect predictions. This is achieved using the **Log Loss** (Binary Cross-Entropy), derived via Maximum Likelihood Estimation (MLE).\n",
+    "\n",
+    "Here is the step-by-step mathematical derivation:\n",
+    "\n",
+    "#### 1. Likelihood of a Single Instance (Bernoulli Distribution)\n",
+    "For a single training instance $(x^{(i)}, y^{(i)})$, the true label is binary: $y^{(i)} \\in \\{0, 1\\}$. If our model predicts the probability $\\hat{y}^{(i)}$, we can express the probability (Likelihood) of observing the true label using the Bernoulli distribution:\n",
+    "\n",
+    "$$P(y^{(i)}|x^{(i)};\\theta) = (\\hat{y}^{(i)})^{y^{(i)}}(1-\\hat{y}^{(i)})^{1-y^{(i)}}$$\n",
+    "\n",
+    "This compact expression works for both possible outcomes:\n",
+    "* If $y^{(i)} = 1$, the probability is $\\hat{y}^{(i)}$.\n",
+    "* If $y^{(i)} = 0$, the probability is $1 - \\hat{y}^{(i)}$.\n",
+    "\n",
+    "#### 2. Joint Likelihood of the Dataset\n",
+    "Assuming that all $m$ training instances are independent, the total likelihood of the model, $L(\\theta)$, is the product of the individual probabilities:\n",
+    "\n",
+    "$$L(\\theta) = \\prod_{i=1}^{m} P(y^{(i)}|x^{(i)};\\theta)$$\n",
+    "\n",
+    "During training, our objective is to find the parameters (weights) that **maximize** this likelihood.\n",
+    "\n",
+    "#### 3. Log-Likelihood\n",
+    "Multiplying thousands of probabilities (numbers between 0 and 1) leads to computational underflow. To fix this, we apply the natural logarithm. This transforms the product into a sum and brings the exponents down as multipliers:\n",
+    "\n",
+    "$$l(\\theta) = \\sum_{i=1}^{m} \\log P(y^{(i)}|x^{(i)};\\theta)$$\n",
+    "$$l(\\theta) = \\sum_{i=1}^{m} \\left[ y^{(i)}\\log(\\hat{y}^{(i)}) + (1-y^{(i)})\\log(1-\\hat{y}^{(i)}) \\right]$$\n",
+    "\n",
+    "#### 4. Negative Log-Likelihood (The Cost Function)\n",
+    "Optimization algorithms like Gradient Descent are designed to **minimize** a cost function rather than maximize it. To convert this into a minimization problem, we multiply the Log-Likelihood by $-1$. Finally, we divide by the total number of samples $m$ to get the average error. \n",
+    "\n",
+    "This gives us the final Log Loss equation:\n",
+    "\n",
+    "$$J(\\theta) = -\\frac{1}{m} \\sum_{i=1}^{m} \\left[ y^{(i)}\\log(\\hat{y}^{(i)}) + (1-y^{(i)})\\log(1-\\hat{y}^{(i)}) \\right]$$"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 8,
@@ -498,7 +551,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 11,
+   "execution_count": null,
    "id": "f1156c02",
    "metadata": {},
    "outputs": [],
@@ -510,9 +563,7 @@
     "\n",
     "def predict(x, theta0, theta1, thresh=0.5):\n",
     "    model = (predictProba >= thresh).astype(int)\n",
-    "    # Returns 1 if probability >= threshold, else 0\n",
-    "\n",
-    "    "
+    "    # Returns 1 if probability >= threshold, else 0"
    ]
   },
   {