📊 Computing Derivatives with a Computation Graph

This notebook explains how to compute derivatives using a computation graph, following the example from Andrew Ng's neural networks course. It breaks down a simple function into smaller steps and shows how to apply the chain rule to calculate how changes in inputs affect the final output.

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🧠 The Function

We are given:

$$J = 3(a + bc)$$

Let’s break it into smaller steps:

• $u = b \times c$
• $v = a + u$
• $J = 3 \times v$

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✅ Forward Pass

Given values:

• $a = 5$
• $b = 3$
• $c = 2$

We compute:

• $u = 3 \times 2 = 6$
• $v = 5 + 6 = 11$
• $J = 3 \times 11 = 33$

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🔁 Backward Pass (Derivatives)

We use the chain rule to compute how much changing each input affects the output $J$.

Step 1: $\frac{dJ}{dv} = 3$

Because:

$$J = 3v \Rightarrow \text{If } v \uparrow 1, J \uparrow 3$$

Step 2: $\frac{dv}{du} = 1$

Because:

$$v = a + u \Rightarrow \text{If } u \uparrow 1, v \uparrow 1$$

Step 3: $\frac{dJ}{du} = \frac{dJ}{dv} \times \frac{dv}{du} = 3 \times 1 = 3$

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📦 Derivatives with Respect to Inputs

1. $\frac{dJ}{da}$

• $v = a + u \Rightarrow \frac{dv}{da} = 1$
• $\frac{dJ}{da} = \frac{dJ}{dv} \times \frac{dv}{da} = 3 \times 1 = 3$

2. $\frac{dJ}{db}$

• $u = b \times c \Rightarrow \frac{du}{db} = c = 2$
• $\frac{dJ}{db} = \frac{dJ}{du} \times \frac{du}{db} = 3 \times 2 = 6$

3. $\frac{dJ}{dc}$

• $u = b \times c \Rightarrow \frac{du}{dc} = b = 3$
• $\frac{dJ}{dc} = \frac{dJ}{du} \times \frac{du}{dc} = 3 \times 3 = 9$

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✅ Final Summary Table

Variable	Derivative
$a$	$\frac{dJ}{da} = 3$
$b$	$\frac{dJ}{db} = 6$
$c$	$\frac{dJ}{dc} = 9$

This tells us how much each input influences the final output $J$ using derivatives and chain rule in a computation graph.