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Exercise 3.1.8

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${(A + Δ A)}^{2} - A^{2} = (A + Δ A) (A + Δ A) - A^{2} = A Δ A + Δ AA + Δ^{2} A$ , divide by $Δ t$ and send $Δ$ to 0, we have $\frac{d A^{2} (t)}{dt} = A \frac{dA (t)}{dt} + \frac{dA (t)}{dt} A$

niuers

2020-03-20 00:00

Exercise 3.1.8

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