Why ∂b is 6?

J = 3a + 3bc \Longrightarrow \frac{\partial J}{\partial b} = 3c, \quad \frac{\partial J}{\partial c} = 3b
And if we follow the lectures way of the chain rule, we’d get the same result, but why did Andrew conclude that \frac{\partial J}{\partial b} = 6, \quad \frac{\partial J}{\partial c} = 9?

What are the values of b and c in the lecture?

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Got it :smile:
b = 3 \Longrightarrow \partial c = 3 \times 3 = 9
c = 2 \Longrightarrow \partial b = 3 \times 2 = 6